Constant factors and polynomials
James Harris
certain math arguments of mine because
P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f)
has that constant factor of f^2.
Normally when considering factorizations, you separate off constant
factors, as otherwise you don't have a unique factorization even with
polynomial factors.
For instance
4(x^2 + 2x + 1) = (2x + 2)(2x + 2) = (x+1)(4x + 4)
along with an infinity of other factorizations, but typically you'd
just have
4(x^2 + 2x + 1) = 4(x+1)(x+1).
Besides all that the expression I use is rather imposing, and it has a
lot of symbols, so I thought I'd remind you of a few things.
1. You *can* look at an actual example with m=1, f=sqrt(2), as then
all that complexity drops away and you have
P(1) = 2x^3 - 3x + 1
which actually does reduce over Q.
Some of you may have realized that you can consider m=1(mod sqrt(2))
to blow apart several assertions made by some posters.
2. A requirement I give is that f be coprime to 3, but letting f=3,
you get that *each* of the a's in the factorization
3^2((m^3 3^4 - 3m^2 3^2 + 3m) x^3 -
3(-1+m3^2 )x u^2 + u^3 3) =
(a_1 x + 3u)(a_2 x + 3u)(a_3 x + 3u)
has a non-unit factor in common with 3, which is a radical factor of
3, and there's no reason to believe it varies with m, or that it cares
if the polynomial is irreducible over Q. That actually destroys
several claims made about using Galois Theory where reducibility over
rationals is an issue.
What I want you to understand is that for trained mathematicians,
these are not issues. However, when it comes to confusing people
about even relatively basic mathematics, who would be better at it
than mathematicians?
They need to confuse you here for *social* reasons.
Notice that with f=3, the constant term P(0) = u^2(3x + 3u) =
3u^2(x+u), so it *still* has a factor that is 3, and that's why I
always have the condition that f be coprime to 3.
Here, however, I'm hoping it'll help to point out why that requirement
is there, and what happens if you ignore it.
Well then, what are some posters trying to convince you about
P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f)?
They're trying to convince you that there is a mathematical limitation
based on reducibility over Q that determines how f^2 can divide
through when you have the factorization
P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
and the first question that should come to you is, how could a
constant factor be constrained by reducibility over rationals?
Now that question is resolvable, but I know the answer is in my favor,
so mathematicians are avoiding even letting you know that IS the
question, and instead those who post work to confuse.
Now given that I know I have a short proof of Fermat's Last Theorem,
and that mathematicians have been avoiding dealing with reality, while
some posters have gotten away with *deliberately* confusing people,
why would I quit talking about my proof of FLT?
If you'd found a short proof of Fermat's Last Theorem, would you quit
talking about it?
James Harris
Toby Meres
You seem to be quite a dedicated amateur mathematician. Good for you. I
was wondering what you do for a living?
Also, Nora Baron has asked that you refute her objections to your proof
in her post 36024859.03080...@posting.google.com
Surely it is in your interest to do so? If you're right about you proof
of FLT, what do you have to lose?
Regards,
Toby
C. Bond
> If you'd found a short proof of Fermat's Last Theorem, would you quit
> talking about it?
No, but if had been refuted I wouldn't keep talking about it, either.
--
There are two things you must never attempt to prove: the unprovable --
and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
The Last Danish Pastry
news:3c65f87.03080...@posting.google.com...
> It occurred to me that some of you may be hampered in understanding
your posts due to the fact that they consist of mindless drivel.
The Ghost In The Machine
<jst...@msn.com>
wrote
on 5 Aug 2003 09:08:00 -0700
<3c65f87.03080...@posting.google.com>:
> It occurred to me that some of you may be hampered in understanding
> certain math arguments of mine because
>
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>
> 3(-1+mf^2 )x u^2 + u^3 f)
>
> has that constant factor of f^2.
>
> Normally when considering factorizations, you separate off constant
> factors, as otherwise you don't have a unique factorization even with
> polynomial factors.
>
> For instance
>
> 4(x^2 + 2x + 1) = (2x + 2)(2x + 2) = (x+1)(4x + 4)
>
> along with an infinity of other factorizations, but typically you'd
> just have
>
> 4(x^2 + 2x + 1) = 4(x+1)(x+1).
I suspect there are a number of ways of dealing with
this issue; I'd probably state that the factorization
would require that all non-trivial polynomials have
coefficients with gcd 1. ("4" is a coefficient of the
trivial polynomial 4 * x^0 and would have to be treated
as a special case, but one can also chop up 4 into its
constituent prime factors if need be.)
>
> Besides all that the expression I use is rather imposing, and it has a
> lot of symbols, so I thought I'd remind you of a few things.
>
> 1. You *can* look at an actual example with m=1, f=sqrt(2), as then
> all that complexity drops away and you have
>
> P(1) = 2x^3 - 3x + 1
>
> which actually does reduce over Q.
P(1) = 2*x^3 - 6*u^2*x + 2*sqrt(2)*u^3
One has to set u to be 1/sqrt(2) as well.
>
> Some of you may have realized that you can consider m=1(mod sqrt(2))
> to blow apart several assertions made by some posters.
P(1 + k(sqrt(2)) =
((2 * sqrt(2)*k^3 + 6*k^2 + 3*sqrt(2)*k + 1)*f^6
+ (-6*k^2 - 6*sqrt(2)*k - 3)*f^4 + (3*sqrt(2)*k + 3)*f^2)*x^3
+ ((-3*sqrt(2)*k - 3)*u^2*f^4 + 3*u^2*f^2)*x + u^3*f^3
for any integer (or, for that matter, non-integer) k.
This is not reducible over Q except when k = 0, even
if f is equal to 2^(1/4).
(This expression courtesy of Pari GP, which may explain its
slight oddity, but I'm not about to bust my brains out
to clean it up except for replacing 2.8284271247... with 2 * sqrt(2),
etc.)
>
> 2. A requirement I give is that f be coprime to 3, but letting f=3,
> you get that *each* of the a's in the factorization
>
> 3^2((m^3 3^4 - 3m^2 3^2 + 3m) x^3 -
>
> 3(-1+m3^2 )x u^2 + u^3 3) =
>
> (a_1 x + 3u)(a_2 x + 3u)(a_3 x + 3u)
>
> has a non-unit factor in common with 3, which is a radical factor of
> 3, and there's no reason to believe it varies with m, or that it cares
> if the polynomial is irreducible over Q. That actually destroys
> several claims made about using Galois Theory where reducibility over
> rationals is an issue.
>
> What I want you to understand is that for trained mathematicians,
> these are not issues. However, when it comes to confusing people
> about even relatively basic mathematics, who would be better at it
> than mathematicians?
Non-mathematicians, in some cases. Training tends to wear a groove
in some people's minds. :-)
>
> They need to confuse you here for *social* reasons.
Mathematics is in part a social science; all sciences are, by
virtue of peer review.
>
> Notice that with f=3, the constant term P(0) = u^2(3x + 3u) =
> 3u^2(x+u), so it *still* has a factor that is 3, and that's why I
> always have the condition that f be coprime to 3.
>
> Here, however, I'm hoping it'll help to point out why that requirement
> is there, and what happens if you ignore it.
>
> Well then, what are some posters trying to convince you about
>
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>
> 3(-1+mf^2 )x u^2 + u^3 f)?
>
> They're trying to convince you that there is a mathematical limitation
> based on reducibility over Q that determines how f^2 can divide
> through when you have the factorization
>
> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
I think there's some confusion. P(m)'s 0 term is in fact u^3*f^3
when multiplied out.
>
> and the first question that should come to you is, how could a
> constant factor be constrained by reducibility over rationals?
>
> Now that question is resolvable, but I know the answer is in my favor,
> so mathematicians are avoiding even letting you know that IS the
> question, and instead those who post work to confuse.
>
> Now given that I know I have a short proof of Fermat's Last Theorem,
> and that mathematicians have been avoiding dealing with reality, while
> some posters have gotten away with *deliberately* confusing people,
> why would I quit talking about my proof of FLT?
>
> If you'd found a short proof of Fermat's Last Theorem, would you quit
> talking about it?
Depends on how many demonstrable errors there were in the proof.
My short perusal through your webpages suggests that you might
want to clarify your thinking and/or show your work a bit more, as
you leap from equation to equation without grinding it out in some
cases. I'd have to look to be more specific at this point.
You might profit by studying Andrew Wiles' proof as well. I don't
know if it's on the Web.
>
>
> James Harris
--
#191, ewi...@earthlink.net
It's still legal to go .sigless.
James Harris
It's easy enough just to separate the 4 to the side as I did above,
but things become more complicated with an expression like
P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f) =
(a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
as can be seen by the *months* of discussion I've gone through, though
now there should be progress as I've nailed down a false assumption
that others must be having, which is the belief the f^2 can divide
from the factors
(a_1 x + uf), (a_2 x + uf), and (a_3 x + uf)
as a function of m.
It's the kind of weird false assumption that can just hang out there
if no one puts it forward directly, and I think that mathematicians
would not make it.
After all, f^2 is a constant factor of P(m), why would it have an m
dependency?
Luckily, I can easily show that it does not for those who get really
stuck on the false assumption.
> >
> > Besides all that the expression I use is rather imposing, and it has a
> > lot of symbols, so I thought I'd remind you of a few things.
> >
> > 1. You *can* look at an actual example with m=1, f=sqrt(2), as then
> > all that complexity drops away and you have
> >
> > P(1) = 2x^3 - 3x + 1
> >
> > which actually does reduce over Q.
>
> P(1) = 2*x^3 - 6*u^2*x + 2*sqrt(2)*u^3
>
> One has to set u to be 1/sqrt(2) as well.
Oh yeah, I left out several steps, like using y=uf. Notice
f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f) =
(m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 -
3(-1+mf^2 )x u^2 f^2 + u^3 f^3.
Now using y=uf, I have
(m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 -
3(-1+mf^2 )xy^2 + y^3.
So, if you factor to get something like
(a_1 x + y)(a_2 x + y)(a_3 x + y)
the a's are independent of y, so I can let y=1, so I have
(m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 -
3(-1+mf^2 )x + 1
and with m=1, f=sqrt(2) that is
2x^3 - 3x + 1.
If you prefer to keep y, you have
2x^3 - 3y^2 + y^3.
If you really *must* keep y=uf, so that you need y=sqrt(2)u, then you
may do so.
The expression is still, of course, reducible over Q.
> >
> > Some of you may have realized that you can consider m=1(mod sqrt(2))
> > to blow apart several assertions made by some posters.
>
> P(1 + k(sqrt(2)) =
> ((2 * sqrt(2)*k^3 + 6*k^2 + 3*sqrt(2)*k + 1)*f^6
> + (-6*k^2 - 6*sqrt(2)*k - 3)*f^4 + (3*sqrt(2)*k + 3)*f^2)*x^3
> + ((-3*sqrt(2)*k - 3)*u^2*f^4 + 3*u^2*f^2)*x + u^3*f^3
>
> for any integer (or, for that matter, non-integer) k.
> This is not reducible over Q except when k = 0, even
> if f is equal to 2^(1/4).
>
> (This expression courtesy of Pari GP, which may explain its
> slight oddity, but I'm not about to bust my brains out
> to clean it up except for replacing 2.8284271247... with 2 * sqrt(2),
> etc.)
And the important point is that it is only reducible for k=0.
That shreds the objections where posters have claimed that
reducibility over Q is actually controlling whther or not two of the
a's have a factor that is f, with
P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f) =
(a_1 x + uf)(a_2 x + uf)(a_3 x + uf).
And it seems that what they were actually depending on was the
possibility of confusion where people falsely assumed that f^2 could
divide out from
(a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
as a *function* or variable dependent on m, despite it being constant.
Luckily that strange and false assumption can be easily refuted by
letting f=3, or letting f have any non unit factor in common with 3,
in case someone thinks that f=3 exactly makes a difference.
I suggest to readers that the math experts never made the strange
assumption, but might have surmised that others could fall prey to it.
> >
> > 2. A requirement I give is that f be coprime to 3, but letting f=3,
> > you get that *each* of the a's in the factorization
> >
> > 3^2((m^3 3^4 - 3m^2 3^2 + 3m) x^3 -
> >
> > 3(-1+m3^2 )x u^2 + u^3 3) =
> >
> > (a_1 x + 3u)(a_2 x + 3u)(a_3 x + 3u)
> >
> > has a non-unit factor in common with 3, which is a radical factor of
> > 3, and there's no reason to believe it varies with m, or that it cares
> > if the polynomial is irreducible over Q. That actually destroys
> > several claims made about using Galois Theory where reducibility over
> > rationals is an issue.
> >
> > What I want you to understand is that for trained mathematicians,
> > these are not issues. However, when it comes to confusing people
> > about even relatively basic mathematics, who would be better at it
> > than mathematicians?
>
> Non-mathematicians, in some cases. Training tends to wear a groove
> in some people's minds. :-)
My work is *basic* algebra. It's hard to believe that discussions
could have gone on for so many months with mathematicians, i.e. math
experts by definition, unaware of the truth.
On the other hand, admitting the truth has a definite social
consequence.
Given the improbability that math experts were in fact lost on strange
and false math assumptions, where they might have seen a clear benefit
to obscuring the truth by various means, it's more reasonable to
suppose that they acted on social motivations.
> >
> > They need to confuse you here for *social* reasons.
>
> Mathematics is in part a social science; all sciences are, by
> virtue of peer review.
That is true. I am, however, not a mathematician. I'm an admitted
discoverer for profit, who has made extraordinary math finds.
Mathematicians may see a social benefit to obscuring my finds from the
world to among other things, preserve their current social structure
and control over mathematics itself.
Power corrupts after all. And consider how much power mathematicians
have now when it comes to saying what is true in mathematics.
> >
> > Notice that with f=3, the constant term P(0) = u^2(3x + 3u) =
> > 3u^2(x+u), so it *still* has a factor that is 3, and that's why I
> > always have the condition that f be coprime to 3.
> >
> > Here, however, I'm hoping it'll help to point out why that requirement
> > is there, and what happens if you ignore it.
> >
> > Well then, what are some posters trying to convince you about
> >
> > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
> >
> > 3(-1+mf^2 )x u^2 + u^3 f)?
> >
> > They're trying to convince you that there is a mathematical limitation
> > based on reducibility over Q that determines how f^2 can divide
> > through when you have the factorization
> >
> > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
> I think there's some confusion. P(m)'s 0 term is in fact u^3*f^3
> when multiplied out.
Nope. Setting m=0, gives P(0) = 3xu^2 + u^3 f = u^2(3x + uf).
What's fascinating about it is that you can see echoes of the
factorization
(a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
as it's clear that at m=0, two and only two of the a's equal 0, as
that's the only way to get that "u^2" in u^2(3x+uf).
It's actually rather fascinating. Which just makes it that much
clearer that mathematicians have been avoiding interesting *math*
instead choosing to focus on obscuring recognition of its validity,
either directly in posts attacking my me or my work, or indirectly by
ignoring my work.
> >
> > and the first question that should come to you is, how could a
> > constant factor be constrained by reducibility over rationals?
> >
> > Now that question is resolvable, but I know the answer is in my favor,
> > so mathematicians are avoiding even letting you know that IS the
> > question, and instead those who post work to confuse.
> >
> > Now given that I know I have a short proof of Fermat's Last Theorem,
> > and that mathematicians have been avoiding dealing with reality, while
> > some posters have gotten away with *deliberately* confusing people,
> > why would I quit talking about my proof of FLT?
> >
> > If you'd found a short proof of Fermat's Last Theorem, would you quit
> > talking about it?
>
> Depends on how many demonstrable errors there were in the proof.
> My short perusal through your webpages suggests that you might
> want to clarify your thinking and/or show your work a bit more, as
> you leap from equation to equation without grinding it out in some
> cases. I'd have to look to be more specific at this point.
Being specific is important, otherwise your comments can't be put into
context.
Giving what I've seen I'm not willing to just be trusting.
And the great thing about mathematics is that I don't have to be.
FYI my website is http://groups.msn.com/AmateurMath
so the proof is out there.
> You might profit by studying Andrew Wiles' proof as well. I don't
> know if it's on the Web.
Why?
James Harris
Nora Baron
Not even close. First, in this case, f factors out of your
polynomial 3 times, not 2 times as in the cases you were
considering (f <> 3). This is a special case of no interest
to you or me. It is irrelevant. The Galois argument does not
apply here. No claims based on that argument are 'destroyed'.
Second, only one of the proofs that you are wrong in the cases
where f <> 3 is dependent on Galois Theory. The other proofs are
based on an elementary theorem from algebraic number theory,
which you have previously accepted. All of the proofs *do*
require irreducibility of P(x)/f^2.
Your statement that
"... there's no reason to believe it varies
with m, or that it cares if the polynomial is irreducible
over Q"
is one of the weakest you have ever put forward. First,
saying " there's no reason to believe it" is one of that lamest
non-proofs I have seen. What you are really saying is,
'*I* can't think of any reason to believe it.' Second, justifying a
conclusion on the basis that "it cares" whether a polynomial is
irreducible - this is truly new methodology. You are basing an
argument on the psychology of some utterly fictional entity. In fact
REDUCIBILITY MATTERS - that is, conclusions regarding irreducible
polynomials are *qualitatively different* from conclusions
regarding reducible ones.
Here is an example:
x^2 -5*x + 6 = 0
has constant term 6 = 2*3. The roots are 2 and 3. One root is
coprime to 3 and the other is coprime to 2.
But
x^2 -3*x + 6 = 0
also has two roots. They are:
r1 = (3 + sqrt(-15))/2 and
r2 = (3 - sqrt(-15))/2.
You can check (by computing a 4th degree polynomial)
that:
NEITHER r1 nor r2 is coprime to 2, and
NEITHER r1 nor r2 is coprime to 3.
Thus: REDUCIBILITY MATTERS: note that:
x^2 - 5*x + 6 is reducible over Q, and
x^2 - 3*x + 6 is irreducible over Q.
It's not a coincidence.
>
>What I want you to understand is that for trained mathematicians,
>these are not issues. However, when it comes to confusing people
>about even relatively basic mathematics, who would be better at it
>than mathematicians?
>
>They need to confuse you here for *social* reasons.
>
This is not true. All we want is that the truth
emerges. Social consequences are irrelevant and
unimportant. You, in contrast, are desperately seeking
social recognition for your "proof". As has happened
in the past, we are dragging you kicking and screaming
into the realization that your "proof" is incorrect.
Although social reasons - recognition, especially - are
paramount for you, you are losing the battle. That is
why you desperately keep starting new threads and you refuse
to answer rigorous arguments which prove you are wrong.
>Notice that with f=3, the constant term P(0) = u^2(3x + 3u) =
>3u^2(x+u), so it *still* has a factor that is 3, and that's why I
>always have the condition that f be coprime to 3.
>
>Here, however, I'm hoping it'll help to point out why that requirement
>is there, and what happens if you ignore it.
>
>Well then, what are some posters trying to convince you about
>
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>
> 3(-1+mf^2 )x u^2 + u^3 f)?
>
>They're trying to convince you that there is a mathematical limitation
>based on reducibility over Q that determines how f^2 can divide
>through when you have the factorization
>
> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
>and the first question that should come to you is, how could a
>constant factor be constrained by reducibility over rationals?
>
See above: REDUCIBILITY MATTERS greatly and is central to
the disagreement here. It is reasonable to ask, as you do, how
reducibility might constrain factoring. It is NOT reasonable to assume
with no hint of proof that it doesn't. Our counterproofs need either
a basic theorem from Galois theory (which *requires* irreducibility), or
a basic theorem from algebraic number theory (which *also requires*
irreducibility). If you really want to understand why, you need to
study the proofs of those theorems. However the quadratic examples I
gave above should be sufficient to show exactly how reducibility can
affect the form of factors of roots.
>Now that question is resolvable, but I know the answer is in my favor,
>so mathematicians are avoiding even letting you know that IS the
>question, and instead those who post work to confuse.
>
>Now given that I know I have a short proof of Fermat's Last Theorem,
>and that mathematicians have been avoiding dealing with reality, while
>some posters have gotten away with *deliberately* confusing people,
>why would I quit talking about my proof of FLT?
>
>If you'd found a short proof of Fermat's Last Theorem, would you quit
>talking about it?
>
No: *** If ***. But if other people had presented rigorous
proofs that I was wrong, I would try very hard to
dismantle those proofs. You have made no effort to do that,
preferring instead to bring up irrelevant issues as in the
present thread. The bottom line here is, whether you like it
or not, reducibility DOES matter. If a ridiculous argument based
on whether "it" (math) "doesn't care" about reducibility is all
you have, you had better give up mathematics.
Nora B.
>
>James Harris
Arturo Magidin
Nora Baron <nora...@hotmail.com> wrote:
[.snip.]
> Your statement that
>
> "... there's no reason to believe it varies
> with m, or that it cares if the polynomial is irreducible
> over Q"
>
>is one of the weakest you have ever put forward. First,
>saying " there's no reason to believe it" is one of that lamest
>non-proofs I have seen. What you are really saying is,
>'*I* can't think of any reason to believe it.' Second, justifying a
>conclusion on the basis that "it cares" whether a polynomial is
>irreducible - this is truly new methodology.
No, it is over a year old. He first started using it to attack the
standard theorem on roots of irreducible polynomials with integer
coefficients. He kept arguing about how the "math" couldn't "care"
whether a polynomial was irreducible or not, and how the roots
couldn't "care", and how it was stupid to think the roots "cared
whether the polynomial is irreducible or not."
It seems just as effective here as it was there, though.
======================================================================
"Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A great
many people are staggered to this extend, that they imagine there
must be the indefinite "something" in the mysterious "all this".
They are brought to the point of suspicion that the mathematicians
ought not to treat "all this" with such undisguised contempt,
at least."
-- "A Budget of Paradoxes", Vol. 2 p. 129 by Augustus de Morgan
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
William Hale
nora...@hotmail.com (Nora Baron) wrote:
> jst...@msn.com (James Harris) wrote in message
news:<3c65f87.03080...@posting.google.com>...
> Your statement that
>
> "... there's no reason to believe it varies
> with m, or that it cares if the polynomial is irreducible
> over Q"
>
> is one of the weakest you have ever put forward. First,
> saying " there's no reason to believe it" is one of that lamest
> non-proofs I have seen. What you are really saying is,
> '*I* can't think of any reason to believe it.' Second, justifying a
> conclusion on the basis that "it cares" whether a polynomial is
> irreducible - this is truly new methodology. You are basing an
> argument on the psychology of some utterly fictional entity. In fact
> REDUCIBILITY MATTERS - that is, conclusions regarding irreducible
> polynomials are *qualitatively different* from conclusions
> regarding reducible ones.
Here's is my point of view of his statement you quoted.
James Harris want to prove a statement about the factors of
a polynomial of a particular general form. Offhand, one
might suppose that the statement is true for some such
polynomials and false for other such polynomials. It could
happen that the statement is false for random particular
polynomials. Or, it may happen that the statement is
false for certain polynomials that satisfy some particular
property (like, being irreducible or having prime degree or
something else). This latter case does not exclude the
statement being false for other random particular polynomials
or being false for some other particular property.
James wants to prove that his statement is true for all
such polynomials. Until I have agreed that his proof
is correct, I can't rule out that his statement "cares"
whether the polynomial is irreducicble or whether its
degree is prime or whether some other property holds
true. In fact, some additional property on such
polynomials might be enough for me to show that
his statement is false (under the additional assumed
property, like irreducible).
If his statement can't care whether the polynomials have
additional properties, then of course his statement that
he is trying to prove would be true. But, I don't know
that a priori.
In fact, if I doubt his statement is true, I might try
to find a counterexample. One way is to randomly
try various polynomials to see if his statement is
false. Of course, this could take a long time. A better
way is to try to analyse the statement into various
cases (eg, irreducible or reducible; or prime degree
not prime degree). Indeed, with the case of irreducible,
I have enough additional properties to show that
his statement is always false for polynomials satisfying
that property (of being irreducible). Note, that this
does not say that the statement is true for all other
such polynomials. For reducible polynomials of such
a form, the statement still may be true always, or
may be false always, or may be sometimes true or
sometimes false. Since I am only interested in finding
at least one counterexample, I need not worry about
analysing the reducible case.
-- Bill Hale
Nora Baron
This is actually a reply to a post by William Hale, who was
replying to my post:
On 06 Aug 2003, William Hale wrote:
>In article <36024859.03080...@posting.google.com>,
>nora...@hotmail.com (Nora Baron) wrote:
>> jst...@msn.com (James Harris) wrote in message
>news:<3c65f87.03080...@posting.google.com>...
>[cut]
>> Your statement that
>>
>> "... there's no reason to believe it varies
>> with m, or that it cares if the polynomial is irreducible
>> over Q"
>>
>> is one of the weakest you have ever put forward. First,
>> saying " there's no reason to believe it" is one of that lamest
>> non-proofs I have seen. What you are really saying is,
>> '*I* can't think of any reason to believe it.' Second, justifying a
>> conclusion on the basis that "it cares" whether a polynomial is
>> irreducible - this is truly new methodology. You are basing an
>> argument on the psychology of some utterly fictional entity. In fact
>> REDUCIBILITY MATTERS - that is, conclusions regarding irreducible
>> polynomials are *qualitatively different* from conclusions
>> regarding reducible ones.
>
>Here's is my point of view of his statement you quoted.
>
>James Harris want to prove a statement about the factors of
>a polynomial of a particular general form. Offhand, one
>might suppose that the statement is true for some such
>polynomials and false for other such polynomials. It could
>happen that the statement is false for random particular
>polynomials. Or, it may happen that the statement is
>false for certain polynomials that satisfy some particular
>property (like, being irreducible or having prime degree or
>something else). This latter case does not exclude the
>statement being false for other random particular polynomials
>or being false for some other particular property.
>
Agreed.
>James wants to prove that his statement is true for all
>such polynomials. Until I have agreed that his proof
>is correct, I can't rule out that his statement "cares"
>whether the polynomial is irreducicble or whether its
>degree is prime or whether some other property holds
>true.
I think when he talks about something that "cares"
about such, he means that "the math" is what "cares" or
"doesn't care". But your interpretation could be right
too.
>In fact, some additional property on such
>polynomials might be enough for me to show that
>his statement is false (under the additional assumed
>property, like irreducible).
>
>If his statement can't care whether the polynomials have
>additional properties, then of course his statement that
>he is trying to prove would be true. But, I don't know
>that a priori.
>
Right.
>In fact, if I doubt his statement is true, I might try
>to find a counterexample. One way is to randomly
>try various polynomials to see if his statement is
>false. Of course, this could take a long time.
It won't. 'Most' polynomials with integer
coefficients are irreducible. Here I am being informal
about the probability statement 'Most'. If, for
example, you randomly generated degree 3 polynomials with
integer coefficients randomly selected between 1 and 1000,
the majority will be irreducible. Computation of factors of
the roots could be tedious however.
>A better
>way is to try to analyse the statement into various
>cases (eg, irreducible or reducible; or prime degree
>not prime degree). Indeed, with the case of irreducible,
>I have enough additional properties to show that
>his statement is always false for polynomials satisfying
>that property (of being irreducible).
Yes - several proofs of that have now been given.
>Note, that this
>does not say that the statement is true for all other
>such polynomials. For reducible polynomials of such
>a form, the statement still may be true always, or
>may be false always, or may be sometimes true or
>sometimes false. Since I am only interested in finding
>at least one counterexample, I need not worry about
>analysing the reducible case.
>
I agree. The case he is interested in for his
FLT "proof" is the irreducible case. Our counterproofs
explicitly mention 'irreducible' as an assumption. He
keeps trying to say that the proofs are wrong because
he can find *reducible* polynomials where the conclusion
is not true. To which the right response is, "Duh?
Why do you think we included irreducible as an assumption?"
Here is the germ of his thinking. He notes that
we say, in the irreducible case, that each root
has algebraic integer factors in common with each
prime divisor of the constant term. For polynomials
of 3rd degree or higher, this is hard to see
explicitly. He cannot follow our arguments and
cannot do the computations for himself, so he
assumes we are lying or wrong. He CAN do the
computations in some reducible cases, and they
do not agree with what we say in the irreducible
cases. He then says, 'How can "the math" care
whether a given polynomial is irreducible?' and
concludes (using faulty intuition) that "it" cannot,
and therefore all the more, we must be lying.
So I gave quadratic examples where he might
possibly be able to do the computations:
one reducible and one irreducible, and
(as expected!) they have very different behavior
with respect to factors of the roots.
He will either ignore this or deny it, or
say the quadratic case is irrelevant (even
though his arguments if valid would apply
equally well there). He knows he is at the point
now where he cannot give an inch on this, or
he loses everything.
He is remarkably creative
in throwing out bogus and irrelevant arguments.
He makes you realize that for most problems,
there are only a very few right solutions, but
infinitely many wrong ones. His mathematical
intuition is actually quite bad, and he comes
up with wrong ideas in an incredibly high
percentage of cases. But eventually he is
going to run out of even remotely plausible
options on this.
Nora B.
>-- Bill Hale
James Harris
Well, it IS the case that for f=sqrt(2) only *two* of the a's have a
factor that is sqrt(2), so your claim that it is otherwise is false.
Readers can easily verify by considering
P(1) = 2x^3 -3xy^2 + y^3, where y=uf,
using f=sqrt(2), m=1, with P(m) given above, as it is reducible.
Then consider that using m=1(mod sqrt(2)) then destroys any claims
that reducibility is the issue, as, for instance, m=3 is NOT
reducible.
Now if "Nora Baron" were an ethical mathematician, there would be no
debate.
But it's not even clear who "Nora Baron" is, as was pointed out by
another poster the name itself is a palindrome.
That is, it's the same reversed as Nora B-aron.
The actual poster may not even be female.
That was hand-waving. In fact, your position requires that people
believe that given
P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f) =
(a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
that some f^2 divides off as some function of m, or variable dependent
on m.
What I noted is the oddity of believing that a constant factor of a
polynomial, which f^2 is, could factor off as a function.
It seems unlikely to me that a math expert would make that mistake.
And in fact in this case it's rather easy to prove that belief is
false by letting f=3, as then ALL of the a's have a factor of 3 that
is 3^{2/3}.
> >
> >What I want you to understand is that for trained mathematicians,
> >these are not issues. However, when it comes to confusing people
> >about even relatively basic mathematics, who would be better at it
> >than mathematicians?
> >
> >They need to confuse you here for *social* reasons.
> >
>
> This is not true. All we want is that the truth
> emerges. Social consequences are irrelevant and
> unimportant. You, in contrast, are desperately seeking
> social recognition for your "proof". As has happened
> in the past, we are dragging you kicking and screaming
> into the realization that your "proof" is incorrect.
Then why are you ignoring the actual math argument that I repeatedly
give?
Besides, why would you care?
There are any number of posters all over Usenet with various notions.
It makes more sense that realizing my work is correct, and fearing the
social consequences of acknowledging the truth, people like yourself
"Nora Baron" bank on your ability to confuse others about the
mathematics.
The more I simplify, the more you desperately work to confuse, which I
think explains your recent postings.
> Although social reasons - recognition, especially - are
> paramount for you, you are losing the battle. That is
> why you desperately keep starting new threads and you refuse
> to answer rigorous arguments which prove you are wrong.
Then why bother continuing to reply?
In actuality, you are lying. You have not given valid math arguments
but instead have given bogus ones.
However, you have relied on using math where you can hide the fact
that it is invalid by various tricks, mostly banking on people not
checking thoroughly.
Faced with your efforts, and the support you get from other poster
helping you to try and obscure the truth, I keep bringing focus back
to the actual math argument I've presented.
>
> >Notice that with f=3, the constant term P(0) = u^2(3x + 3u) =
> >3u^2(x+u), so it *still* has a factor that is 3, and that's why I
> >always have the condition that f be coprime to 3.
> >
> >Here, however, I'm hoping it'll help to point out why that requirement
> >is there, and what happens if you ignore it.
> >
> >Well then, what are some posters trying to convince you about
> >
> > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
> >
> > 3(-1+mf^2 )x u^2 + u^3 f)?
> >
> >They're trying to convince you that there is a mathematical limitation
> >based on reducibility over Q that determines how f^2 can divide
> >through when you have the factorization
> >
> > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
> >
> >and the first question that should come to you is, how could a
> >constant factor be constrained by reducibility over rationals?
> >
>
> See above: REDUCIBILITY MATTERS greatly and is central to
> the disagreement here. It is reasonable to ask, as you do, how
> reducibility might constrain factoring. It is NOT reasonable to assume
> with no hint of proof that it doesn't. Our counterproofs need either
> a basic theorem from Galois theory (which *requires* irreducibility), or
> a basic theorem from algebraic number theory (which *also requires*
> irreducibility). If you really want to understand why, you need to
> study the proofs of those theorems. However the quadratic examples I
> gave above should be sufficient to show exactly how reducibility can
> affect the form of factors of roots.
Yet I can give f=sqrt(2), with m=1, and get a reducible polynomial,
which you already made a false statement about in this post, where
only two of the a's have a factor that is sqrt(2).
Then I can simply use m=1(mod sqrt(2)) for a family of results where
most are irreducible.
Now your claim would require that suddenly, against mathematical
logic, the single one of the a's that doesn't have a non-unit factor
in common with sqrt(2) gets one.
Faced with the destruction of your argument, you side-stepped it,
which diligent readers can see at the start of this post.
> >Now that question is resolvable, but I know the answer is in my favor,
> >so mathematicians are avoiding even letting you know that IS the
> >question, and instead those who post work to confuse.
> >
> >Now given that I know I have a short proof of Fermat's Last Theorem,
> >and that mathematicians have been avoiding dealing with reality, while
> >some posters have gotten away with *deliberately* confusing people,
> >why would I quit talking about my proof of FLT?
> >
> >If you'd found a short proof of Fermat's Last Theorem, would you quit
> >talking about it?
> >
>
>
> No: *** If ***. But if other people had presented rigorous
> proofs that I was wrong, I would try very hard to
> dismantle those proofs. You have made no effort to do that,
> preferring instead to bring up irrelevant issues as in the
> present thread. The bottom line here is, whether you like it
> or not, reducibility DOES matter. If a ridiculous argument based
> on whether "it" (math) "doesn't care" about reducibility is all
> you have, you had better give up mathematics.
>
> Nora B.
So now even though I have a proof of my own, you claim that I should
have to work to show falsity in alternate "proofs" given by people who
have proven a need to deceive, who have a long record of deception,
who may be actual mathematicians, i.e. math experts, with a
willingness to use their knowledge to confuse others?
Here readers should consider that "Nora Baron" is probably a pseudonym
that the poster believes is clever because it's a palindrome, who may
not even be a woman.
They can also consider the clear dodge of an issue brought up against
the poster's claims with f=sqrt(2), m=1.
But best of all, readers can consider that *mathematics* gives them
their best way to figure out what's going on, which is to consider the
argument "Nora Baron" keeps working to obscure.
Consider
P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f.
Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization
P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
where w_1 w_2 w_3 = f, and
b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),
and at m=0
P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
so two of the b's must equal 0, which means
P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)
which is
P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)
proving that w_1 w_2 must equal 1, as f is coprime to 3 from before,
which leaves b_3 = 3.
Essentially objections now come down to claiming that the w's are
dependent on m, but consider that w_1 w_2 = 1, when m=0, here where f
is coprime to 3.
But that was an arbitrary choice *I* made, so let f=3.
Now w_1 w_2 = 3^{2/3} as long as m is coprime to 3, WITHOUT REGARD TO
m.
So those posters who try to convince you that the w's are actually
dependent on m, like being functions of m, must now also convince you
that the w's make a decision, first looking to see if f=3 or have some
non-unit factor in common with 3, and THEN they decide if they're
dependent on m.
People can waffle trying to figure out who they are, but mathematics
is logical, which is why I've emphasized that posters are acting on
*social* not mathematical reasons.
James Harris
C. Bond
[snip]
> Here readers should consider that "Nora Baron" is probably a pseudonym
> that the poster believes is clever because it's a palindrome, who may
> not even be a woman.
How is that a math issue? Perhaps "James Harris" is an anagram for "M. Harie Ass, Jr." and you have cleverly
rearranged it.
KRamsay
(James Harris) writes:
|What I noted is the oddity of believing that a constant factor of a
|polynomial, which f^2 is, could factor off as a function.
|
|It seems unlikely to me that a math expert would make that mistake.
You've only guessed that "the way f^2 divides off" is constant. The
mistake a mathematician wouldn't make is to just keep saying
"how odd" it would be if their guess were wrong, and pretend that that's
a "logical argument". Accusing others of "oddly" believing otherwise
is just posturing. It's not up to others to prove it's false; it's up to you
to prove that it's true.
Here's a related example. The two roots r1(m) and r2(m) of x^2+x+2m=0
multiply together to give 2m, which is divisible by 2. At m=0 one of them
is divisible by 2 and the other one is coprime to 2. But there's no
simple pattern in their common factors with 2:
m=1: r1=(-1+sqrt(-7))/2, r2=(-1-sqrt(-7))/2. 2m=2=r1*r2. But that means
that r1 and r2 are themselves the algebraic integer factors of 2 dividing
r1 and r2. Neither r1 nor r2 is divisible by 2 in the algebraic integers.
m=2: r1=(-1+sqrt(-15))/2, r2=(-1-sqrt(-15))/2. The algebraic integer
factors that r1 and r2 have in common with 2 are sqrt(r1) and sqrt(r2).
Neither r1 nor r2 is divisible by 2.
m=-1: r1=1, r2=-2. The algebraic integer factors that r1 and r2 have
in common with 2 are again 1 and 2.
Keith Ramsay
Wayne Brown
> Here readers should consider that "Nora Baron" is probably a pseudonym
> that the poster believes is clever because it's a palindrome, who may
> not even be a woman.
Well, if so, then it was too clever for you; you didn't catch on until
someone else pointed it out.
Anyway, idiot, her identity is irrelevant. All that counts is that
she's right and you're wrong.
--
Wayne Brown | "When your tail's in a crack, you improvise
fwb...@bellsouth.net | if you're good enough. Otherwise you give
| your pelt to the trapper."
"e^(i*pi) = -1" -- Euler | -- John Myers Myers, "Silverlock"
James Harris
> In article <36024859.03080...@posting.google.com>,
> Nora Baron <nora...@hotmail.com> wrote:
>
> [.snip.]
>
> > Your statement that
> >
> > "... there's no reason to believe it varies
> > with m, or that it cares if the polynomial is irreducible
> > over Q"
> >
> >is one of the weakest you have ever put forward. First,
> >saying " there's no reason to believe it" is one of that lamest
> >non-proofs I have seen. What you are really saying is,
> >'*I* can't think of any reason to believe it.' Second, justifying a
> >conclusion on the basis that "it cares" whether a polynomial is
> >irreducible - this is truly new methodology.
>
>
> No, it is over a year old. He first started using it to attack the
> standard theorem on roots of irreducible polynomials with integer
> coefficients. He kept arguing about how the "math" couldn't "care"
> whether a polynomial was irreducible or not, and how the roots
> couldn't "care", and how it was stupid to think the roots "cared
> whether the polynomial is irreducible or not."
I've been pointing out now that a constant factor of a polynomial does
not divide off as a function, while for some time posters like Magidin
have gotten away with objections that depend on saying that m=0 is a
special case, forcing an m dependency on how f^2 divides off.
> It seems just as effective here as it was there, though.
People can say all kinds of things.
Mathematics is not about just trusting what people say.
Here's what follows from the math:
Arturo Magidin
It's interesting that you ignore any post I make with math in it, and
when I make an offhand comment, you jump in to make non sequitur
claims and assertions that usually have little base in reality, or
little sense as written.
Your post was a perfect example. Since turnaround is fair play, I
guess, I will also ignore everything you posted and instead make some
comments. My ignoring your post here is not an admission of any kind,
since I have addressed your multiple errors many times before.
Some time ago I guessed at what was behind your intuition. Based on
many flawed arguments you have given from time to time, and senseless
statements like the above, my guess is that you are thinking in terms
of polynomials.
If P(x) is a polynomial with coefficients in a commutative ring R,
then if r=s (mod a) for r,s,a in R, then one can easily prove that
P(r)=P(s) (mod A). From this it follows of course that P(a)=P(0) (mod
A), and from this we obtain the result
PROPOSITION. Let P(x) be a polynomial with coefficients in the
commutative ring R. For all a in R, P(a) is coprime to a if and only
if P(0) is coprime to a.
This is essentially the argument you have been trying to use. Of
course, your functions are not polynomials, but you still imagine them
as ->essentially<- polynomials, perhaps polynomials in rational powers
of x instead of integer powers, perhaps involving radicals and so
on. Once before you already tried to argue that a=b (mod r) implies
sqrt(a)=sqrt(b) (mod r), and never really acknowledged that this is
simply false.
So, what are you doing? You imagine your coefficients as functions of
m, and you imagine that these "functions" are really a lot like
polynomials. So you evaluate at 0 to figure out what the "constant term"
is, and thus deduce that since this "constant term" is coprime to certain
values of m when evaluated at 0, it must also be coprime when
evaluated at any value coprime to that constant term, just like a
polynomial would.
Of course, the error is pretty dismal. It ignores the fact that
congruences do not behave well with general algebraic functions, they
behave terribly with radicals, AND ignores that your functions are
really piece-wise defined. For p=3, you get general formulas for the
coefficients (in some instances at least) from the Cardano formulas,
but m=0 is a degenerate case in which you cannot use those formulas
but must instead switch to specific values. In these cases, the
argument you are probably basing your intuition on is bollocks.
Bill Dubuque
>
> If P(x) is a polynomial with coefficients in a commutative ring R,
> then if r=s (mod a) for r,s,a in R, then one can easily prove that
> P(r)=P(s) (mod A). From this it follows of course that P(a)=P(0) (mod
> A), and from this we obtain the result
>
> PROPOSITION. Let P(x) be a polynomial with coefficients in the
> commutative ring R. For all a in R, P(a) is coprime to a if and only
> if P(0) is coprime to a.
i.e. (a ,P(a)) = (a ,P(0)) i.e. a==0 => P(a)==P(0)
or (a-b,P(a)) = (a-b,P(b)) i.e. a==b => P(a)==P(b)
so a-b | P(a)-P(b) e.g. 10-1 | P(10)-P(1) "Casting Out Nines"
These are prototypical examples encountered when one first studies
congruences and (evaluation) homomorphisms on abstract rings. That
polynomial evaluation is a ring hom is already known in high-school,
as are applications such as Castings Out Nines. Many of my prior
posts discuss at length related ideas, e.g. start with this one:
http://groups.google.com/groups?selm=y8zu1dgidp9.fsf%40nestle.ai.mit.edu
-Bill Dubuque
James Harris
Hmmm...so I give the math argument, you delete it out, and then give
weird excuses.
> Your post was a perfect example. Since turnaround is fair play, I
> guess, I will also ignore everything you posted and instead make some
> comments. My ignoring your post here is not an admission of any kind,
> since I have addressed your multiple errors many times before.
Ok, so I give the math argument, you delete it out, give weird
excuses, and then claim it's "fair play", and, oh yeah, you claim that
you've addressed "multiple errors" before.
Gist of it is, you're telling readers to trust you.
I say trust the mathematics, which I will give, once again.
Well, I don't think so, but rather than just toss out *opinions* I'm
going to give the actual mathematics and readers can work it out for
themselves.
> Of course, the error is pretty dismal. It ignores the fact that
> congruences do not behave well with general algebraic functions, they
> behave terribly with radicals, AND ignores that your functions are
> really piece-wise defined. For p=3, you get general formulas for the
> coefficients (in some instances at least) from the Cardano formulas,
> but m=0 is a degenerate case in which you cannot use those formulas
> but must instead switch to specific values. In these cases, the
> argument you are probably basing your intuition on is bollocks.
Well you said "probably basing" in that paragraph, and there is no
need to guess, as I'll give the mathematics.
Consider, in the ring of algebraic integers,
P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f).
Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization
P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
where w_1 w_2 w_3 = f, and
b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),
and at m=0
P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
so two of the b's must equal 0, which means
P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)
which is
P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)
proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves
b_3 = 3.
Essentially objections to how f^2 divides off now come down to
claiming that the w's are functions of m, but consider that w_1 w_2 =
1, when m=0, if f is coprime to 3.
But that was an arbitrary choice, so let f=3.
Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m.
And in fact, all of the w's are constant values for ALL m when f=3.
Therefore, the factorization is
P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f =
(b_1 x + u)(b_2 x + u)(b_3 x + uf)
where you'll notice that the b's are algebraic integers with m=1,
f=sqrt(2), but that's a special case as generally they are not, which
shows a problem with the ring of algebraic integers.
James Harris
Nora Baron
In your main applications, as in your "proof" of FLT,
f is an integer. I assumed that here.
> Readers can easily verify by considering
>
> P(1) = 2x^3 -3xy^2 + y^3, where y=uf,
>
> using f=sqrt(2), m=1, with P(m) given above, as it is reducible.
>
Relatively minor point: if f = sqrt(2), then f^3 = 2*sqrt(2),
which is not rational, so y = uf gives an irrational constant
term, y^3, for your polynomial. It makes no sense to discuss reducibility
(over the rationals) of a polynomial unless it has rational
coefficients. This one does not when f = sqrt(2).
> Then consider that using m=1(mod sqrt(2)) then destroys any claims
> that reducibility is the issue, as, for instance, m=3 is NOT
> reducible.
You've lost me here. See above note regarding reducibility of
this polynomial when it does not even have rational coefficients.
As for m = 3, I assume you intended to say that the polynomial
is irreducible over Q when m = 3. Again this doesn't make any sense
when the polynomial does not even have rational coefficients.
A side question: are you saying that m = 3 is congruent to 1
modulo sqrt(2) ???? Would you care to explain that?
That was absolutely NOT handwaving. There is nothing in what
I said but cold, hard computation. I gave two examples: one,
a reducible polynomial, for which the roots are coprime to one
or the other factor of the constant term; the other, an
IRREDUCIBLE polynomial, for which NEITHER root is coprime to
ANY factor of the constant term. These are not isolated
examples; these are the general rule. The point: IRREDUCIBILITY
MATTERS. Somehow, in your terms, "the math knows" or
"the math cares", whether you like it or not.
And you blow it off by calling it hand-waving! Tell me
how I can be more explicit!
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>
> 3(-1+mf^2 )x u^2 + u^3 f) =
>
> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
> that some f^2 divides off as some function of m, or variable dependent
> on m.
>
YES! YES! YES! YOU GOT IT!
> What I noted is the oddity of believing that a constant factor of a
> polynomial, which f^2 is, could factor off as a function.
>
Odd or not, that is how it works. The way it factors depends
on its reducibility or irreducibility. That should not seem strange.
Lots of theorems in algebraic number theory depend on reducibility or
irreducibility.
> It seems unlikely to me that a math expert would make that mistake.
>
Ask other "math experts". Anyone you like.
> And in fact in this case it's rather easy to prove that belief is
> false by letting f=3, as then ALL of the a's have a factor of 3 that
> is 3^{2/3}.
>
In that case, unlike the one in which you are interested,
P(x) / f^2 is not coprime to f. Since your own arguments depend
on such coprimeness, this is an irrelevant example.
> > >
> > >What I want you to understand is that for trained mathematicians,
> > >these are not issues. However, when it comes to confusing people
> > >about even relatively basic mathematics, who would be better at it
> > >than mathematicians?
> > >
> > >They need to confuse you here for *social* reasons.
> > >
> >
> > This is not true. All we want is that the truth
> > emerges. Social consequences are irrelevant and
> > unimportant. You, in contrast, are desperately seeking
> > social recognition for your "proof". As has happened
> > in the past, we are dragging you kicking and screaming
> > into the realization that your "proof" is incorrect.
>
> Then why are you ignoring the actual math argument that I repeatedly
> give?
>
No one here has paid more detailed attention to your argument
than I have. I have looked at it on its own terms. This is
more than I should have done. I and others have given explicit,
rigorous proofs that your main claims are incorrect. What this
means is, there is no particular reason to do your work for you
and identify where you have made your mistake. I have nevertheless
pinpointed your error many times, and that is what you are now
trying to deal with: how you justify going from a factorization
when m = 0 to a factorization when m <> 0. Nothing I have seen
so far from you on this hangs together logically.
> Besides, why would you care?
>
> There are any number of posters all over Usenet with various notions.
>
For sure. There are loonies, mystics, jokers, liars, etc., of
all sorts out there. Most of them do not pretend to do mathematics and
they cannot be proven right or wrong. You however try to present
logical arguments. You say things that annoy me because they are
provably false, and you brag about it. My instinct is to not let
you get away with it.
> It makes more sense that realizing my work is correct, and fearing the
> social consequences of acknowledging the truth, people like yourself
> "Nora Baron" bank on your ability to confuse others about the
> mathematics.
>
I very definitely do not want to confuse anyone. I try to write
simply and clearly and I put in most of the details. I do not
want to mislead anyone. As for social consequences, if I honestly
thought you had a valid proof, I would defend it as strongly as I
attack what you have at present.
> The more I simplify, the more you desperately work to confuse, which I
> think explains your recent postings.
>
That is one possible explanation. Another is that I believe what
I am saying and my main goal is to see that the truth prevails, that I
do not like someone claiming false results when they can easily be
debunked.
> > Although social reasons - recognition, especially - are
> > paramount for you, you are losing the battle. That is
> > why you desperately keep starting new threads and you refuse
> > to answer rigorous arguments which prove you are wrong.
>
> Then why bother continuing to reply?
>
I am hopeful that you will eventually understand what is wrong
with your argument.
> In actuality, you are lying. You have not given valid math arguments
> but instead have given bogus ones.
>
ABSOLUTELY NOT!
> However, you have relied on using math where you can hide the fact
> that it is invalid by various tricks, mostly banking on people not
> checking thoroughly.
>
People: check all you want, and everything you want. I abhor
'tricks'. I like simple straightforward stuff, and that is what
I have tried to present.
If you think I am tricking you, point out what it is you don't
understand. The examples I gave above of quadratic polynomials -
was that a trick? Why do you think so? You called it handwaving.
Did you feel I was trying to trick you? If so, ASK FOR AN
EXPLANATION of any step you didn't get.
> Faced with your efforts, and the support you get from other poster
> helping you to try and obscure the truth, I keep bringing focus back
> to the actual math argument I've presented.
>
Great.
See above. Since f = sqrt(2) and the constant term is (u*f)^3,
where u is an integer, the constant term is not rational. So it
makes no sense to discuss reducibility over the rationals for this
polynomial.
> Then I can simply use m=1(mod sqrt(2)) for a family of results where
> most are irreducible.
>
> Now your claim would require that suddenly, against mathematical
> logic, the single one of the a's that doesn't have a non-unit factor
> in common with sqrt(2) gets one.
>
> Faced with the destruction of your argument, you side-stepped it,
> which diligent readers can see at the start of this post.
>
Please first resolve the problem with f^3 being irrational
before you go any farther with this. Perhaps you have a definition
of irreducible (over the rationals) that is at variance with the
usual one.
> > >Now that question is resolvable, but I know the answer is in my favor,
> > >so mathematicians are avoiding even letting you know that IS the
> > >question, and instead those who post work to confuse.
> > >
> > >Now given that I know I have a short proof of Fermat's Last Theorem,
> > >and that mathematicians have been avoiding dealing with reality, while
> > >some posters have gotten away with *deliberately* confusing people,
> > >why would I quit talking about my proof of FLT?
> > >
> > >If you'd found a short proof of Fermat's Last Theorem, would you quit
> > >talking about it?
> > >
> >
> >
> > No: *** If ***. But if other people had presented rigorous
> > proofs that I was wrong, I would try very hard to
> > dismantle those proofs. You have made no effort to do that,
> > preferring instead to bring up irrelevant issues as in the
> > present thread. The bottom line here is, whether you like it
> > or not, reducibility DOES matter. If a ridiculous argument based
> > on whether "it" (math) "doesn't care" about reducibility is all
> > you have, you had better give up mathematics.
> >
> > Nora B.
>
> So now even though I have a proof of my own,
No, dammit, I said *** IF ***. You don't have a proof.
You have written down many "proofs" over the years. You
have admitted over and over again that what you thought
was a proof was in fact wrong. You have made stupid
mistakes in various arguments even in the last few days.
Why you NOW think you have suddenly become infallible baffles
the hell out of me.
> you claim that I should
> have to work to show falsity in alternate "proofs" given by people who
> have proven a need to deceive, who have a long record of deception,
> who may be actual mathematicians, i.e. math experts, with a
> willingness to use their knowledge to confuse others?
>
If I were you I would try to find errors in the counterarguments.
If you won't listen to us, maybe it would be reasonable for us to
post a warning to editors of math journals, saying there are strong
counterarguments to claims you are making in a paper you are trying
to publish; specifying what those counterarguments are; and noting
that, although you have had multiple opportunities, you have not
found errors in any of them. You are either (1) refusing to do so,
or (2) unable to do so. In either case, it is dishonest of you to try to
publish a result when there are real, rigorous, unanswered objections
which prove it is false. Should we notify editors of this?
> Here readers should consider that "Nora Baron" is probably a pseudonym
> that the poster believes is clever because it's a palindrome, who may
> not even be a woman.
>
> They can also consider the clear dodge of an issue brought up against
> the poster's claims with f=sqrt(2), m=1.
>
See above. It should not be news that sqrt(2) is not a
rational number.
> But best of all, readers can consider that *mathematics* gives them
> their best way to figure out what's going on, which is to consider the
> argument "Nora Baron" keeps working to obscure.
>
> Consider
>
> P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>
> 3(-1+mf^2 )x u^2 + u^3 f.
>
>
> Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization
>
> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
>
> where w_1 w_2 w_3 = f, and
>
> b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),
>
> and at m=0
>
> P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
>
> so two of the b's must equal 0, which means
>
> P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)
>
> which is
>
> P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)
>
> proving that w_1 w_2 must equal 1, as f is coprime to 3 from before,
> which leaves b_3 = 3.
>
... right. And every bit of what you have just said is
predicated on the condition:
***** m = 0 *****.
> Essentially objections now come down to claiming that the w's are
> dependent on m,
Exactly! Prove they're not!
> but consider that w_1 w_2 = 1, when m=0, here where f
> is coprime to 3.
>
> But that was an arbitrary choice *I* made, so let f=3.
>
In that case, P(x) / f^2 is not coprime to f, so your
own argument breaks down. It is a special case of no
interest to either you or me. Why don't you look in detail
at the cases f = 5, f = 7, f = 11, ... ?
> Now w_1 w_2 = 3^{2/3} as long as m is coprime to 3, WITHOUT REGARD TO
> m.
>
Really ? I note that you say that u*w3 = u*f = u*3,
that is, w3 = 3, and so if w1*w2 = 3^{2/3}, then
w1*w2*w3 = 3^{5/3}.
But u^3*w1*w2*w3 is supposed to equal the constant term,
which here is u^3 * f = u^3 * 3. Let u = 1 to make this
simpler if you want. Anyway,
3 is not equal to 3^{5/3}.
You want to think about that some more?
> So those posters who try to convince you that the w's are actually
> dependent on m, like being functions of m, must now also convince you
> that the w's make a decision, first looking to see if f=3 or have some
> non-unit factor in common with 3, and THEN they decide if they're
> dependent on m.
>
This is a silly way of saying it, but in a sense it is right.
The form of the factorization DOES depend on reducibility of the
polynomial. Trying to anthropomorphize the w's does not constitute
a proof that they behave the way you want them to. Got that? YOUR
SILLY LITTLE ARGUMENT ABOVE IS NOT A PROOF! If you can present a REAL proof,
do so. The burden of doing this is on you. If you think the w's
are NOT dependent on m, you need to PROVE IT. Bear in mind, however,
that in the cases where P(x) / f^2 is primitive and irreducible,
there are rigorous arguments that say you can't do it. You might
learn something by trying to understand and answer those arguments.
Nora B.
Wayne Brown
> If you won't listen to us, maybe it would be reasonable for us to
> post a warning to editors of math journals, saying there are strong
> counterarguments to claims you are making in a paper you are trying
> to publish; specifying what those counterarguments are; and noting
> that, although you have had multiple opportunities, you have not
> found errors in any of them. You are either (1) refusing to do so,
> or (2) unable to do so. In either case, it is dishonest of you to try to
> publish a result when there are real, rigorous, unanswered objections
> which prove it is false. Should we notify editors of this?
I think that's an excellent idea. It's unlikely that a reputable journal
would be fooled by James' pseudo-math, but it could save some editors
and/or reviewers from wasting too much time on it.
James Harris
> jst...@msn.com (James Harris) wrote in message news:<3c65f87.03080...@posting.google.com>...
> > nora...@hotmail.com (Nora Baron) wrote in message news:<36024859.03080...@posting.google.com>...
<deleted>
> > > Not even close. First, in this case, f factors out of your
> > > polynomial 3 times, not 2 times as in the cases you were
> > > considering (f <> 3). This is a special case of no interest
> > > to you or me. It is irrelevant. The Galois argument does not
> > > apply here. No claims based on that argument are 'destroyed'.
> > > Second, only one of the proofs that you are wrong in the cases
> > > where f <> 3 is dependent on Galois Theory. The other proofs are
> > > based on an elementary theorem from algebraic number theory,
> > > which you have previously accepted. All of the proofs *do*
> > > require irreducibility of P(x)/f^2.
> >
> > Well, it IS the case that for f=sqrt(2) only *two* of the a's have a
> > factor that is sqrt(2), so your claim that it is otherwise is false.
> >
>
> In your main applications, as in your "proof" of FLT,
> f is an integer. I assumed that here.
Your assumption is irrelevant to that fact, as you have tried to
confuse people by working to convince that m=0 is a special case.
You have in the past succeeded, but now I'm showing, rather easily and
directly that you were indeed misleading them.
> > Readers can easily verify by considering
> >
> > P(1) = 2x^3 -3xy^2 + y^3, where y=uf,
> >
> > using f=sqrt(2), m=1, with P(m) given above, as it is reducible.
> >
>
> Relatively minor point: if f = sqrt(2), then f^3 = 2*sqrt(2),
> which is not rational, so y = uf gives an irrational constant
> term, y^3, for your polynomial. It makes no sense to discuss reducibility
> (over the rationals) of a polynomial unless it has rational
> coefficients. This one does not when f = sqrt(2).
As usual you work to mislead, and here it's by distraction.
While it is true that given
2x^3 -3xy^2 + y^3
with y=uf, where f=sqrt(2), you end up with coefficients that are not
rational.
It is also true that 2x^3 -3xy^2 + y^3 factors over Q.
That's important because I show that you are wrong with the emphasis
on m=0 as a special case by showing m=1, where people can see that my
argument correctly gives ONLY two of the a's as having a factor that
is sqrt(2), while the remaining one is coprime to 2.
>
> > Then consider that using m=1(mod sqrt(2)) then destroys any claims
> > that reducibility is the issue, as, for instance, m=3 is NOT
> > reducible.
>
>
> You've lost me here. See above note regarding reducibility of
> this polynomial when it does not even have rational coefficients.
> As for m = 3, I assume you intended to say that the polynomial
> is irreducible over Q when m = 3. Again this doesn't make any sense
> when the polynomial does not even have rational coefficients.
Hmmm...so you're claiming it's irreducible over Q? Then that refutes
you directly as your own claim is that reducibility is the issue.
Since you may back off if you see that you can't confuse people on
that issue, I'll point out that using m=1(mod f) with f=sqrt(2), takes
away the possibility that the factors of f, move, which refutes you as
well.
The *mathematical* basis for your claims is thus destroyed.
> A side question: are you saying that m = 3 is congruent to 1
> modulo sqrt(2) ???? Would you care to explain that?
3 = 1(mod sqrt(2))
Understand?
> >
> > Now if "Nora Baron" were an ethical mathematician, there would be no
> > debate.
> >
> > But it's not even clear who "Nora Baron" is, as was pointed out by
> > another poster the name itself is a palindrome.
> >
> > That is, it's the same reversed as Nora B-aron.
> >
> > The actual poster may not even be female.
<deleted>
> > >
> > > Thus: REDUCIBILITY MATTERS: note that:
> > >
> > > x^2 - 5*x + 6 is reducible over Q, and
> > >
> > > x^2 - 3*x + 6 is irreducible over Q.
> > >
> > > It's not a coincidence.
> >
> > That was hand-waving. In fact, your position requires that people
> > believe that given
> >
>
> That was absolutely NOT handwaving. There is nothing in what
> I said but cold, hard computation. I gave two examples: one,
> a reducible polynomial, for which the roots are coprime to one
> or the other factor of the constant term; the other, an
> IRREDUCIBLE polynomial, for which NEITHER root is coprime to
> ANY factor of the constant term. These are not isolated
> examples; these are the general rule. The point: IRREDUCIBILITY
> MATTERS. Somehow, in your terms, "the math knows" or
> "the math cares", whether you like it or not.
Your quadratics are irrelevant to the case as hand, and in fact, it's
fascinating that you believe that you've proven some rule, with two
quadratics.
> And you blow it off by calling it hand-waving! Tell me
> how I can be more explicit!
You can stick to the line of mathematical logic, rather than trying to
distract.
>
> > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
> >
> > 3(-1+mf^2 )x u^2 + u^3 f) =
> >
> > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
> >
> > that some f^2 divides off as some function of m, or variable dependent
> > on m.
> >
>
>
> YES! YES! YES! YOU GOT IT!
<deleted>
Previous to that I said that your position requires that people
believe that given
P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f) =
(a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
that some f^2 divides off as some function of m, or variable dependent
on m.
Splitting that sentence up with your own comments and then posting as
if I'd said something else is clearly dishonest.
James Harris
The Last Danish Pastry
news:zlOYa.3095$Pa4....@fe03.atl2.webusenet.com...
> In alt.math.undergrad Nora Baron <nora...@hotmail.com> wrote:
>
> > If you won't listen to us, maybe it would be reasonable for us to
> > post a warning to editors of math journals, saying there are strong
> > counterarguments to claims you are making in a paper you are trying
> > to publish; specifying what those counterarguments are; and noting
> > that, although you have had multiple opportunities, you have not
> > found errors in any of them. You are either (1) refusing to do so,
> > or (2) unable to do so. In either case, it is dishonest of you to try
to
> > publish a result when there are real, rigorous, unanswered objections
> > which prove it is false. Should we notify editors of this?
>
> I think that's an excellent idea. It's unlikely that a reputable journal
> would be fooled by James' pseudo-math, but it could save some editors
> and/or reviewers from wasting too much time on it.
James Harris is but one of many math cranks in this world. I imagine that
journal editors can recognise their nonsense rather quickly. I doubt that
there is any point in warning journal editors about James.
--
Clive Tooth
http://www.clivetooth.dk
Randy Poe
>
> > If you won't listen to us, maybe it would be reasonable for us to
> > post a warning to editors of math journals, saying there are strong
> > counterarguments to claims you are making in a paper you are trying
> > to publish; specifying what those counterarguments are; and noting
> > that, although you have had multiple opportunities, you have not
> > found errors in any of them. You are either (1) refusing to do so,
> > or (2) unable to do so. In either case, it is dishonest of you to try to
> > publish a result when there are real, rigorous, unanswered objections
> > which prove it is false. Should we notify editors of this?
>
> I think that's an excellent idea. It's unlikely that a reputable journal
> would be fooled by James' pseudo-math, but it could save some editors
> and/or reviewers from wasting too much time on it.
I think it's a terrible idea. You're just opening yourself up
to legitimate claims of conspiring. I would hate to think that
such a letter could influence the review process, causing an
editor to prejudge a paper. If it could, just imagine the
possibilities for abuse. A paper should stand or fall on its
merits, and a reviewer should judge a paper on its content.
Trust the system. They're already used to getting incoherent
documents.
- Randy
Wayne Brown
I thought Nora was talking about something other than a letter saying
"This guy's a crank." It was my assumption that by "specifying what those
counterarguments are" she meant that the editor would be provided with
valid mathematical reasons why James' paper is incorrect. The editor
(or reviewer) would be able to read both James' submission and the
counterarguments and make his/her own decision.
It would be a little like this: James shows up at a newspaper with a
picture he's taken of a flying saucer. The editor says, "Now that's
interesting, because I have another picture here that someone sent me.
See this? Here you are in your back yard, with a camera, some fishing
line, and a pie plate that looks a *lot* like that flying saucer of
yours..."
I assume Nora was not seriously proposing to do this, though, and I'm
certainly not qualified to do so.
Arturo Magidin
Weird is in the eye of the beholder. Oh, by the way, what you give is
not a "math argument", it is wishful thinking and sophistry and
bluff.
For instance, I could easily accuse you of cowardice and dishonesty
for ignoring the legion of posts where I have proven you are wrong.
Here's one, for example, from the recent past:
http://groups.google.com/groups?selm=bgefp6%244p3%241%40agate.berkeley.edu
(posted August 1, 2003; you never replied to that one)
-- Begin Insert --
>> That simple lemma not only anchors my paper, and my proof of
Fermat's
>> Last Theorem, it can be used in lots of places. Now I'll use it
with
>> the roots of
>>
>> y^3 + 3y - 2
>
>I find its roots fascinating.
>
>> and the three roots are
>>
>> y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3}
>>
>> y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 -
>>
>> (-1-sqrt(-3))/2(1+sqrt(2))^{1/3}
>>
>> y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 -
>>
>> (-1+sqrt(-3))/2(1+sqrt(2))^{1/3}
>
>As simple as they are you can't just take the roots and *look* at
them
>to tell whether or not any one of them is coprime to 2. It is just
>impossible.
No, it is not impossible. It is amazingly trivial, as Keith Ramsay
pointed out right after you posted them:
NONE OF THEM ARE COPRIME TO 2. THEY ARE ALL DIVISORS OF 2:
y_1*(y_1^2+3) = 2
y_2*(y_2^2+3) = 2
y_3*(y_3^2+3) = 2.
Surely you agree with all three of those expressions, since they are
all roots of y^3 + 3y - 2.
So they are all divisors of 2, and since none of them are units, they
are none of them coprime to 2. Simple, nay, trivial.
-- End Insert --
Same thing happened with
http://groups.google.com/groups?selm=bgemnn%249i8%241%40agate.berkeley.edu
also posted on August 1st, which you also ignored.
Or
http://groups.google.com/groups?selm=benboa%24135c%241%40agate.berkeley.edu
posted on July 11, which included the four extensive treatments of
your false claims, and which you studiously have ignored for months now.
So much for dealing with the math.
>> Your post was a perfect example. Since turnaround is fair play, I
>> guess, I will also ignore everything you posted and instead make some
>> comments. My ignoring your post here is not an admission of any kind,
>> since I have addressed your multiple errors many times before.
>
>Ok, so I give the math argument, you delete it out, give weird
>excuses, and then claim it's "fair play", and, oh yeah, you claim that
>you've addressed "multiple errors" before.
Yeah, just like you, except I bothered to reply instead of sticking my
fingers in my ears and humming real loud.
>Gist of it is, you're telling readers to trust you.
No. I'm telling you that you should answer my mathematical posts, not
the posts where I am commenting on your verifiable behavior to
others. The readers do not need to trust me, they can go and check my
posts and verify for themselves whether what I say is correct or not.
There is not trust involved. It doesn't matter how often you try to
put your lies into my mouth.
[.snip.]
Your argument is still incorrect. You continue to argue that specific
cases will generalize to arbitrary cases by personal
incredulity. That's not a proof. That's wishful thinking.
As to why you are wrong:
Proof that James's claims are incorrect:
THEOREM. Let f(x) be any primitive polynomial of degree n>0 with
integer coefficients, and assume that f(x) is irreducible over
Q. Assume further that
f(x) = (a1*x+b1)...(an*x+bn)
is a factorization of f(x) into linear terms with algebraic integer
coefficients. Then for every integer m and every positive rational
number q,
(i) a1 is coprime to m^q if and only if all ai are coprime to m^q
(in the ring of all algebraic integers).
(ii) b1 is coprime to m^q if and only if all ai are coprime to m^q.
(in the ring of all algebraic integers).
In particular, if p is a rational prime which divides b1*...*bn, then
none of the bi are coprime to p. If r is a rational prime which
divides a1*...*an, then none of the ai are copriem to r.
PROOF. Since f(x) is irreducible over Q, a1,...,an are conjugate up to
units, and b1,...,bn are conjugate up to units: for each pair i,j of
integers between 1 and n, there exists an automorphism of the ring of
all algebraic integers, call it g, and an algebraic integer unit u,
such that g(ai)=u*aj and g(bi)=u*bj.
To see this, let K be the smallest normal extension of Q which includes
a1,...,an,b1,...,bn. Since bi/ai is a root of f(x), and the same for
bj/aj, we know there exists an automorphism of k which sends ai/bi to
aj/bj. By a slight abuse of language, call it g. Then g(ai)*bj =
aj*g(bi). Since f(x) is primitive, we know that ai and bi are coprime
in the ring of all algebraic integers. Therefore, since aj divides
g(ai)*bj and aj is coprime to bj, it follows that aj|g(ai); likewise,
bj|g(bi). Applying the inverse of g to bj/aj, we obtain that
bi|g^{-1}(bj) and ai|g^{-1}(aj); all these divisibiility conditions
occur in the ring of integers of K. Applying g to these divisibility
conditions we obtain that g(bi)|bj, and g(ai)|aj. Combining with
bj|g(bi) and aj|g(ai), we obtain that u*aj=g(ai) for some algebraic
integer unit u, and v*bj=g(bi) for some algebraic integer unit v; both
units lie in the ring of integers of K. In addition, since bj/aj =
g(ai)/g(bi), it follows that u=v, proving the assertion in the first
paragraph of the proof.
It is a standard result that the automorphism g of K may be extended
to the field of all algebraic numbers, and that this will restrict to
an automorphism of the ring of all algebraic numbers.
Note that for m and q as in the statemente, g(m^q) = v*m^q for some
algebraic integer unit v. To see this, write q=r/s with r,s positive
integers, gcd(r,s)=1. Then m^q is a root of x^s-m^r = 0, and therefore
so must g(m^q); but all the roots of x^s-m^r are of the form v*m^q
with v an s-th root of unity, giving the stated equality.
Clearly, if all ai are coprime to m^q, then so is a1. Conversely,
suppose that a1 is coprime to m^q. Let i be any integer, i=2,...,n,
and assume that x is a common algebraic integer factor of ai and
m^q. Let g be the automorphism sending ai to u*a1 for some unit u, and
let g(m^q) = v*m^q, with v a unit.
Since x divides ai and divides m^q, it follows that g(x) divides g(a1)
and g(m^q). But g(a) = u*a1, and g(m^q)=v*m^q; therefore, g(x) must
be a unit, since a1 and m^q are coprime, and hence so are u*a1 and
v*m^q. The same argument holds for the bj. This proves clauses (i) and
(ii) of the theorem.
For the final two statements, simply note that if x and y are
algebraic integers which are coprime to the algebraic integer z, then
so is x*y: we know that there are algebraic integers r,s,t,u such that
rx+sz = 1, and such that ty+uz = 1. Multiplying the two expressions
together, we get
1 = (rt)xy + ruxz + styz + suz^2
= (rt)(xy) + (rux + sty + suz)*z
so xy and z are coprime. Since a1*...*an is not coprime to r, it
follows that not all ai are coprime to r. From clause (i), it follows
that none of the ai are coprime to r. The same arguments holds for the
bj.
Note the importance of the irreducibility and primitivity of f(x). QED
Nora Baron
>nora...@hotmail.com (Nora Baron) wrote in message news:<36024859.03080...@posting.google.com>...
>> jst...@msn.com (James Harris) wrote in message news:<3c65f87.03080...@posting.google.com>...
>> > nora...@hotmail.com (Nora Baron) wrote in message news:<36024859.03080...@posting.google.com>...
>
><deleted>
>
>> > > Not even close. First, in this case, f factors out of your
>> > > polynomial 3 times, not 2 times as in the cases you were
>> > > considering (f <> 3). This is a special case of no interest
>> > > to you or me. It is irrelevant. The Galois argument does not
>> > > apply here. No claims based on that argument are 'destroyed'.
>> > > Second, only one of the proofs that you are wrong in the cases
>> > > where f <> 3 is dependent on Galois Theory. The other proofs are
>> > > based on an elementary theorem from algebraic number theory,
>> > > which you have previously accepted. All of the proofs *do*
>> > > require irreducibility of P(x)/f^2.
>> >
>> > Well, it IS the case that for f=sqrt(2) only *two* of the a's have a
>> > factor that is sqrt(2), so your claim that it is otherwise is false.
>> >
>>
>> In your main applications, as in your "proof" of FLT,
>> f is an integer. I assumed that here.
>
>Your assumption is irrelevant to that fact, as you have tried to
>confuse people by working to convince that m=0 is a special case.
>
1. My assumption is clearly more relevant than assuming f = sqrt(2).
Your applications in either "Advanced Polynomial
Factorization" or your "proof" of FLT both involve assuming f
is a prime *integer*.
2. Since I honestly believe m = 0 *IS* a special case, it would
be less than honest of me to pretend otherwise. I am not the
least bit interested in confusing people.
>You have in the past succeeded, but now I'm showing, rather easily and
>directly that you were indeed misleading them.
>
>> > Readers can easily verify by considering
>> >
>> > P(1) = 2x^3 -3xy^2 + y^3, where y=uf,
>> >
>> > using f=sqrt(2), m=1, with P(m) given above, as it is reducible.
>> >
>>
>> Relatively minor point: if f = sqrt(2), then f^3 = 2*sqrt(2),
>> which is not rational, so y = uf gives an irrational constant
>> term, y^3, for your polynomial. It makes no sense to discuss reducibility
>> (over the rationals) of a polynomial unless it has rational
>> coefficients. This one does not when f = sqrt(2).
>
>As usual you work to mislead, and here it's by distraction.
>
>While it is true that given
>
> 2x^3 -3xy^2 + y^3
>
>with y=uf, where f=sqrt(2), you end up with coefficients that are not
>rational.
>
Right. So it cannot factor over the rationals.
>It is also true that 2x^3 -3xy^2 + y^3 factors over Q.
>
That would mean by definition that this polynomial factors into two
polynomials of lower degree, each of which has rational
coefficients. If y = sqrt(2), the constant term is irrational.
Show that factorization. I really want to see it. Lots of other
people will be interested as well.
>That's important because I show that you are wrong with the emphasis
>on m=0 as a special case by showing m=1, where people can see that my
>argument correctly gives ONLY two of the a's as having a factor that
>is sqrt(2), while the remaining one is coprime to 2.
>
>>
>> > Then consider that using m=1(mod sqrt(2)) then destroys any claims
>> > that reducibility is the issue, as, for instance, m=3 is NOT
>> > reducible.
>>
>>
>> You've lost me here. See above note regarding reducibility of
>> this polynomial when it does not even have rational coefficients.
>> As for m = 3, I assume you intended to say that the polynomial
>> is irreducible over Q when m = 3. Again this doesn't make any sense
>> when the polynomial does not even have rational coefficients.
>
>Hmmm...so you're claiming it's irreducible over Q? Then that refutes
>you directly as your own claim is that reducibility is the issue.
>
It is not even a polynomial to which the term "irreducible over Q"
should be applied. It does not have rational coefficients. If you
insist on describing it as "reducible over Q", then it must equal a
product of lower-degree polynomials with *rational* coefficients.
Clearly that is impossible. Prove me wrong.
>Since you may back off if you see that you can't confuse people on
>that issue, I'll point out that using m=1(mod f) with f=sqrt(2), takes
>away the possibility that the factors of f, move, which refutes you as
>well.
>
>The *mathematical* basis for your claims is thus destroyed.
>
My proof showing that your conclusion is wrong assumes f = 5
and m = 1. Your claimed main results deal with f a *prime integer*.
Clearly f = sqrt(2) is not exactly in the ballpark.
>> A side question: are you saying that m = 3 is congruent to 1
>> modulo sqrt(2) ???? Would you care to explain that?
>
> 3 = 1(mod sqrt(2))
>
>Understand?
>
You must be thinking that the ring in which you are doing
the arithmetic includes the square root of 2. Thus it could
be the ring of algebraic integers, or it could be Z[sqrt(2)].
Thus 3 = 1 + k*sqrt(2) where k = sqrt(2) would suffice. Is
that what you meant?
They were examples. No, I do not claim that examples prove rules.
I thought in this case that examples might convince you that things
were not so simple as you want them to be. I anticipated that you
would say that quadratics are irrelevant and said so. However if
your principles of factorization were correct, that the form of the
factorization is not dependent on reducibility, then the second of
the two example quadratics should not exist: *both* roots share
algebraic integer factors with both integer factors of the constant
term.
>> And you blow it off by calling it hand-waving! Tell me
>> how I can be more explicit!
>
>You can stick to the line of mathematical logic, rather than trying to
>distract.
>
>>
>> > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>> >
>> > 3(-1+mf^2 )x u^2 + u^3 f) =
>> >
>> > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>> >
>> > that some f^2 divides off as some function of m, or variable dependent
>> > on m.
>> >
>>
>>
>> YES! YES! YES! YOU GOT IT!
>
><deleted>
>
>Previous to that I said that your position requires that people
>believe that given
>
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>
> 3(-1+mf^2 )x u^2 + u^3 f) =
>
> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
>that some f^2 divides off as some function of m, or variable dependent
>on m.
>
Yes. How f^2 factors out of the linear terms depends on
whether the polynomial is irreducible, and that in turn
can depend on m.
You say the factorization has a constant form regardless of m. You
said this earlier in this thread about such factorizations:
"So now those posters who try to convince you that the w's
are actually dependent on m, like being functions of m,
must now also convince you that the w's make a decision,
first looking to see if f = 3 or have some non-unit
factor in common with 3, and THEN they decide if
they're dependent on m."
Such a cute way of putting it. Those whimsical little w's, making
decisions on the basis of coefficients of polynomials. Sounds
absurd, doesn't it?
And THIS is what you call a mathematical proof ??!! This is a new
low-water mark. Rather than producing an actual proof, you dismiss
the idea by *personifying* the w's. No wonder you are
cross-posting to alt.fiction.
The fact is, the form of the factorization DOES differ depending
on whether or not the polynomial is reducible. That is exactly what
was illustrated by the quadratic examples, which you called
"handwaving".
Let's get back to real math.
You have said, applying your methods in "Advanced Polynomial
Factorization", that if you factor the polynomial
P(x) = 65*x^3 - 12*x + 1
in the form
P(x) = (a1*x + 1)*(a2*x + 1)*(a3*x + 1),
where a1, a2, and a3 are algebraic integers, then two of
the a's are divisible by sqrt(5) in the algebraic
integers. Right?
Say a1 is divisible by sqrt(5). Let a1 = sqrt(5) * c1,
where c1 is an algebraic integer.
Note that -1/a1 is a root of P(x). Therefore
P(-1/(sqrt(5)*c1)) = 0.
This implies
-65*(1/(5*sqrt(5)*c1^3)) - 12*(-1/(sqrt(5)*c1)) + 1 = 0.
Multiply through by 1/(5*sqrt*5)*c1^3). You get
-65 + 12*5*c1^2 + 5*sqrt(5)*c1^3 = 0.
Divide out 5, move things around:
sqrt(5)*c1^3 = -12*c1^2 + 13.
Square both sides:
5*c1^6 = 144*c1^4 - 312*c1^2 + 169.
Rewrite this as
5*c1^6 - 144*c1^4 + 312*c1^2 - 169 = 0.
Use your favorite piece of software to show that this is
a non-monic and ***irreducible*** polynomial in c1.
Then apply a well-known theorem from algebraic number
theory:
THEOREM: If r is a root of a non-monic polynomial
with integer coefficients, ***irreducible*** over
the rationals, the r cannot be an algebraic integer.
and conclude that c1 cannot be an algebraic integer.
This contradicts your claim that at least one of a1,
a2, or a3 is divisible (in the algebraic integers) by
sqrt(5). Therefore that claim is wrong. Your "proof"
of it has an error.
Have I tricked you somewhere in the above argument?
Have I left out anything? Have I pulled the wool over
your eyes? Have I used phrases like "c1 cannot be an
algebraic integer because it decides not to be one in
this situation"?
Note that irreducibility is an absolutely key assumption
in the Theorem. That is why it plays such a part here.
IRREDUCIBILITY MATTERS.
>Splitting that sentence up with your own comments and then posting as
>if I'd said something else is clearly dishonest.
>
I did NOT post as if you said something else. I did not in any
way misquote you or change your words. I totally agree with you
that my proof implies that you must conclude that the way f^2
factors out of your polynomial DOES depend on whether that
polynomial is reducible or irreducible. I don't understand at all
why that should be surprising. If you believe that the form of the
factorization DOES NOT depend on reducibility, then you should be
able to give an actual proof, and not some bullshit argument about
the w's not having the capacity to "decide" how they should factor
depending on the coefficients. That is *NOT* mathematics; that is
more like Aesop's Fables. The w's unfortunately do not factor the
way you want them to. Similarly, above, contrary to your wishes, NONE
of the a's above are divisible by sqrt(5).
I think the situation is a little bit like quantum theory.
Things at the atomic level do not behave like objects in the macro
world. The double-slit experiment and the 'entanglement'
experiments prove that. The behavior is *unintutitive*. We don't
see these phenomena in everyday life because the scale on which
they take place is very small. Here, polynomials factor in
familiar ways, just like in high school factoring problems, when
they are reducible. But when they are irreducible, they factor
differently, *unintuitively*. It is hard to see this happen
directly because in the irreducible case, the computations are
formidable, even in quadratics. However denying in the face of
overwhelming evidence - proof, actually - that they DO happen is
just silly. What you have done is turn your back on that evidence
and pretend that it does not exist or that we are lying about it or
embedding tricks that you cannot detect.
We're not.
You have several choices here. One is to try to give a real
mathematical proof that the w's factor the way you want them to.
Another is to try to find an error in one of my proofs [note however
that continuing to give irrelevant examples for which the hypotheses
are not satisfied is pointless]. One is to concede my arguments and
those of others, and start over. One is to continue calling me a
liar and try to convince someone - anyone, perhaps - that your
argument makes sense and that, e.g., my proof above is a trick.
You deleted much of my message. You had accused me of lying, to
which I said (and say now) ABSOLUTELY NOT. You deleted the
additional discussion of your referring to the polynomial with an
irrational constant term as reducible over the rationals. You
deleted the part where you said
w1*w2 = 3^{2/3},
and I noted that w3 = 3, and that this was inconsistent with the fact
that
w1*w2*w3 = 3.
I wonder why you omitted that, among other things. You have since
repeated this same error in another post.
Nora B.
>
>James Harris
Arturo Magidin
Nitpicking...
You must also include the hypothesis that the polynomial is
primitive. Since nonzero constants are units in Q[x], they are not
considered nontrivial factors, so the hypothesis must be explicitly
included.
David C. Ullrich
wrote:
That's what I assumed she meant as well, probably because that's
what she said. It's a terrible idea, for the reasons Randy suggests,
and also not necessary, for reasons he suggests.
>It would be a little like this: James shows up at a newspaper with a
>picture he's taken of a flying saucer. The editor says, "Now that's
>interesting, because I have another picture here that someone sent me.
>See this? Here you are in your back yard, with a camera, some fishing
>line, and a pie plate that looks a *lot* like that flying saucer of
>yours..."
>
>I assume Nora was not seriously proposing to do this, though, and I'm
>certainly not qualified to do so.
************************
David C. Ullrich
James Harris
> In article <36024859.03080...@posting.google.com>,
> Nora Baron <nora...@hotmail.com> wrote:
>
> Nitpicking...
>
> > Let's get back to real math.
> >
> > You have said, applying your methods in "Advanced Polynomial
> >Factorization", that if you factor the polynomial
> >
> > P(x) = 65*x^3 - 12*x + 1
> >
> >in the form
> >
> > P(x) = (a1*x + 1)*(a2*x + 1)*(a3*x + 1),
> >
> >where a1, a2, and a3 are algebraic integers, then two of
> >the a's are divisible by sqrt(5) in the algebraic
> >integers. Right?
Nope.
That is correct. And it is true that c1 cannot be an algebraic
integer.
That has not been under debate.
It's also NOT under debate as to whether or not given
65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)
any a's exist, within the ring of algebraic integers, such that
sqrt(5) is a factor of them in that ring.
The problem is that neither a_1, a_2, nor a_3 have ANY non-unit
factors in common with 5 in the ring of algebraic integers.
James Harris
Nora Baron
I see that the current version of APF does not make this
claim, though I believe you have said exactly that in the past.
Here is what the current version of APF *does* claim:
Let f = prime > 3, m = integer coprime to f, v = -1 + m*f^2,
and u = integer coprime to f, and
P(x) = (v^3 + 1)*x^3 + 3*v*x*u^2*f^2 + u^3*f^3.
Then P(x)/f^2 may be factored in the form
[1] P(x)/f^2 = (a1*x + u)*(a2*x + u)*(a3*x + u*f),
where a1, a2, and a3 are algebraic integers.
--------------------------------------------------------
Let f = 5, m = 1, u = 1. Then
P(x)/f^2 = 553*x^3 + 72*x + 5.
If this is factored in form [1], it will look like
P(x)/f^2 = (a1*x + 1)*(a2*x + 1)*(a3*x + 5).
This means that -1/a1 is a root of
553*x^3 + 72*x + 5 = 0.
That is,
553*(-1/a1^3) + 72*(-1/a1) + 5 = 0, or
5*a1^3 -72*a1^2 - 553 = 0.
But this last expression is a non-monic, irreducible,
primitive polynomial in a1. Therefore a1 cannot
be an algebraic integer. Therefore the conclusion
of APF is false.
> The problem is that neither a_1, a_2, nor a_3 have ANY non-unit
> factors in common with 5 in the ring of algebraic integers.
>
That would certainly be a problem, given that their product
is 65.
So it looks like both of us arrive at a contradiction. We
draw different conclusions from it, apparently. I conclude
that your claim is false and that therefore there is necessarily
an error in your proof [and I have described where that error
is and what it is at length].
You conclude that there is something wrong with the ring of
algebraic integers, perhaps that it is "incomplete". What
that means is not clear, at least not to me. It could mean
that the a.i.'s do not really form a ring - perhaps that they
are not closed under addition and multiplication or that the
distributive law does not hold, etc.. This however is a very
old theorem and is not in doubt.
The main question here is, if you arrive at a contradiction,
why do you assume the problem must be somewhere other than in your
own "proof"? Why, in view of your atrocious track record over
8 years, do you now assume that you are infallible?
Have you had a message from God, or what?
Nora B.
>
> James Harris
Wayne Brown
> On Fri, 08 Aug 2003 19:07:25 GMT, Wayne Brown <fwb...@bellsouth.net>
> wrote:
>>I thought Nora was talking about something other than a letter saying
>>"This guy's a crank." It was my assumption that by "specifying what those
>>counterarguments are" she meant that the editor would be provided with
>>valid mathematical reasons why James' paper is incorrect. The editor
>>(or reviewer) would be able to read both James' submission and the
>>counterarguments and make his/her own decision.
> That's what I assumed she meant as well, probably because that's
> what she said. It's a terrible idea, for the reasons Randy suggests,
> and also not necessary, for reasons he suggests.
OK, fine. It just seems to me that as long as the information is correct,
then the source, or how it was obtained, or even the motives of the
person providing it are unimportant. But I've had no experience with
the review process and am happy to defer to those who do.
Arturo Magidin
[.snip.]
>> >theory:
>> >
>> > THEOREM: If r is a root of a non-monic polynomial
>> > with integer coefficients, ***irreducible*** over
>> > the rationals, the r cannot be an algebraic integer.
>> >
>> >and conclude that c1 cannot be an algebraic integer.
>>
>> You must also include the hypothesis that the polynomial is
>> primitive. Since nonzero constants are units in Q[x], they are not
>> considered nontrivial factors, so the hypothesis must be explicitly
>> included.
>
>That is correct. And it is true that c1 cannot be an algebraic
>integer.
>
>That has not been under debate.
Which is why I said I was "nitpicking": pointing out a minor error
that is well understood.
>It's also NOT under debate as to whether or not given
>
> 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)
>
>any a's exist, within the ring of algebraic integers, such that
>sqrt(5) is a factor of them in that ring.
This is rather confused. You have a "not" at the beginning, a "whether
or not" after that, and a qualifier "any" for the a's. It's pretty
close to nonsense. What you are really saying, presumably, is:
"Given a1, a2, a3 algebraic integers such that
65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)
[as a polynomial identity], then none of a_1,a_2,a_3 are multiples
(in the ring of algebraic integers) of sqrt(5)."
This is also true, and has been established.
>The problem is that neither a_1, a_2, nor a_3 have ANY non-unit
>factors in common with 5 in the ring of algebraic integers.
And that's false. I am pretty sure that Dale produced explicit common
factors; but in any case, your claim here is certainly false, since
their product is not coprime to 65.
Lemma. Let R be the ring of all algebraic integers, and let a, b, c be
any elements of R. If a and b are coprime to c, then a*b is coprime to
c.
Proof. We use the characterization of coprime valid for commutative
rings with 1: a and b are coprime in R if and only if there exist x
and y in R such that ax+by = 1.
Since a and c are coprime by assumption, there exist n and m in R such
that an+cm = 1. Since b and c are coprime by assumption, there exist r
and s in R such that br+cs = 1.
Multiplying both together, we have
1 = (an+cm)(br+cs)
= abrn + acns + cbmr + c^2*ms
= ab(rn) + c(ans + bmr + cms).
Let x = rn, y = ans+bmr+cms. Then x and y are algebraic integrs, and
ab*x + c*y = 1. Therefore, ab and y are coprime. QED
So, assume you were correct and neither a_1, a_2, nor a_3 have ANY
non-unit factors in common with 5 in the ring of algebraic
integers. Then, by the lemma, neither does a1*a_2; and applying the
lemma again, neither does a_1*a_2*a_3. But a_1*a_1*a_3 = 65, which
clearly has 5 as a nonunit common factor with 5. This contradicts the
assumption that none of a_1, a_2, a_3 have common non-unit factors
with 5 in the ring of algebraic integers. Therefore, your assertion is
false.
The Ghost In The Machine
<jst...@msn.com>
wrote
on 6 Aug 2003 11:55:43 -0700
<3c65f87.03080...@posting.google.com>:
> The Ghost In The Machine <ew...@sirius.athghost7038suus.net> wrote in message news:<nh7701-...@lexi2.athghost7038suus.net>...
>> In sci.math, James Harris
>> <jst...@msn.com>
>> wrote
>> on 5 Aug 2003 09:08:00 -0700
>> <3c65f87.03080...@posting.google.com>:
>> > It occurred to me that some of you may be hampered in understanding
>> > certain math arguments of mine because
>> >
>> >
>> > 3(-1+mf^2 )x u^2 + u^3 f)
>> >
>> >
>> > Normally when considering factorizations, you separate off constant
>> > factors, as otherwise you don't have a unique factorization even with
>> > polynomial factors.
>> >
>> > For instance
>> >
>> > 4(x^2 + 2x + 1) = (2x + 2)(2x + 2) = (x+1)(4x + 4)
>> >
>> > along with an infinity of other factorizations, but typically you'd
>> > just have
>> >
>> > 4(x^2 + 2x + 1) = 4(x+1)(x+1).
>>
>> this issue; I'd probably state that the factorization
>> would require that all non-trivial polynomials have
>> coefficients with gcd 1. ("4" is a coefficient of the
>> trivial polynomial 4 * x^0 and would have to be treated
>> as a special case, but one can also chop up 4 into its
>> constituent prime factors if need be.)
>
> It's easy enough just to separate the 4 to the side as I did above,
> but things become more complicated with an expression like
>
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>
> 3(-1+mf^2 )x u^2 + u^3 f) =
>
> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
> now there should be progress as I've nailed down a false assumption
> that others must be having, which is the belief the f^2 can divide
> from the factors
>
> (a_1 x + uf), (a_2 x + uf), and (a_3 x + uf)
>
> as a function of m.
But f^2 *can* divide from those factors. If f is prime,
one merely needs to have either exactly one a be divisible
by f^2 (in which case u and f have to relate somehow)
or exactly two a's be divisible by f. If f is a nonprime
additional possibilities ensue, depending on u. I'd have
to work out the gloppy details.
This subproblem gets a little complicated, at first blush, but
it's not unmanageable.
>
> It's the kind of weird false assumption that can just hang out there
> if no one puts it forward directly, and I think that mathematicians
> would not make it.
>
> After all, f^2 is a constant factor of P(m), why would it have an m
> dependency?
>
> Luckily, I can easily show that it does not for those who get really
> stuck on the false assumption.
>
>> >
>> > Besides all that the expression I use is rather imposing, and it has a
>> > lot of symbols, so I thought I'd remind you of a few things.
>> >
>> > 1. You *can* look at an actual example with m=1, f=sqrt(2), as then
>> > all that complexity drops away and you have
>> >
>> > P(1) = 2x^3 - 3x + 1
>> >
>> > which actually does reduce over Q.
>>
>> P(1) = 2*x^3 - 6*u^2*x + 2*sqrt(2)*u^3
>>
>> One has to set u to be 1/sqrt(2) as well.
>
> Oh yeah, I left out several steps, like using y=uf. Notice
>
> f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>
> 3(-1+mf^2 )x u^2 + u^3 f) =
>
> (m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 -
>
> 3(-1+mf^2 )x u^2 f^2 + u^3 f^3.
>
> Now using y=uf, I have
>
> (m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 -
>
> 3(-1+mf^2 )xy^2 + y^3.
>
> So, if you factor to get something like
>
> (a_1 x + y)(a_2 x + y)(a_3 x + y)
>
> the a's are independent of y, so I can let y=1, so I have
>
>
> (m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 -
>
> 3(-1+mf^2 )x + 1
>
> and with m=1, f=sqrt(2) that is
>
> 2x^3 - 3x + 1.
>
> If you prefer to keep y, you have
>
> 2x^3 - 3y^2 + y^3.
>
> If you really *must* keep y=uf, so that you need y=sqrt(2)u, then you
> may do so.
>
> The expression is still, of course, reducible over Q.
That it is.
>
>
>> >
>> > Some of you may have realized that you can consider m=1(mod sqrt(2))
>> > to blow apart several assertions made by some posters.
>>
>> P(1 + k(sqrt(2)) =
>> ((2 * sqrt(2)*k^3 + 6*k^2 + 3*sqrt(2)*k + 1)*f^6
>> + (-6*k^2 - 6*sqrt(2)*k - 3)*f^4 + (3*sqrt(2)*k + 3)*f^2)*x^3
>> + ((-3*sqrt(2)*k - 3)*u^2*f^4 + 3*u^2*f^2)*x + u^3*f^3
>>
>> for any integer (or, for that matter, non-integer) k.
>> This is not reducible over Q except when k = 0, even
>> if f is equal to 2^(1/4).
>>
>> (This expression courtesy of Pari GP, which may explain its
>> slight oddity, but I'm not about to bust my brains out
>> to clean it up except for replacing 2.8284271247... with 2 * sqrt(2),
>> etc.)
>
> And the important point is that it is only reducible for k=0.
>
> That shreds the objections where posters have claimed that
> reducibility over Q is actually controlling whther or not two of the
> a's have a factor that is f, with
>
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>
> 3(-1+mf^2 )x u^2 + u^3 f) =
>
> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf).
>
> And it seems that what they were actually depending on was the
> possibility of confusion where people falsely assumed that f^2 could
> divide out from
>
> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
> as a *function* or variable dependent on m, despite it being constant.
>
> Luckily that strange and false assumption can be easily refuted by
> letting f=3, or letting f have any non unit factor in common with 3,
> in case someone thinks that f=3 exactly makes a difference.
It doesn't.
If one equates your P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
the a's by necessity have to have certain properties, especially
if one's assuming the a's are all rational. However, since you've
defined P(m) as a product of f^2 with something else, one can
always compute, say, Q(m) = P(m) / f^2, fairly trivially. The
divisor is not dependent on m. (Whether this is useful is
not clear.)
>
> I suggest to readers that the math experts never made the strange
> assumption, but might have surmised that others could fall prey to it.
>
>> >
>> > 2. A requirement I give is that f be coprime to 3, but letting f=3,
>> > you get that *each* of the a's in the factorization
>> >
>> > 3^2((m^3 3^4 - 3m^2 3^2 + 3m) x^3 -
>> >
>> > 3(-1+m3^2 )x u^2 + u^3 3) =
>> >
>> > (a_1 x + 3u)(a_2 x + 3u)(a_3 x + 3u)
>> >
>> > has a non-unit factor in common with 3, which is a radical factor of
>> > 3, and there's no reason to believe it varies with m, or that it cares
>> > if the polynomial is irreducible over Q. That actually destroys
>> > several claims made about using Galois Theory where reducibility over
>> > rationals is an issue.
>> >
>> > What I want you to understand is that for trained mathematicians,
>> > these are not issues. However, when it comes to confusing people
>> > about even relatively basic mathematics, who would be better at it
>> > than mathematicians?
>>
>> Non-mathematicians, in some cases. Training tends to wear a groove
>> in some people's minds. :-)
>
> My work is *basic* algebra. It's hard to believe that discussions
> could have gone on for so many months with mathematicians, i.e. math
> experts by definition, unaware of the truth.
>
> On the other hand, admitting the truth has a definite social
> consequence.
>
> Given the improbability that math experts were in fact lost on strange
> and false math assumptions, where they might have seen a clear benefit
> to obscuring the truth by various means, it's more reasonable to
> suppose that they acted on social motivations.
>
>> >
>> > They need to confuse you here for *social* reasons.
>>
>> Mathematics is in part a social science; all sciences are, by
>> virtue of peer review.
>
> That is true. I am, however, not a mathematician. I'm an admitted
> discoverer for profit, who has made extraordinary math finds.
>
> Mathematicians may see a social benefit to obscuring my finds from the
> world to among other things, preserve their current social structure
> and control over mathematics itself.
>
> Power corrupts after all. And consider how much power mathematicians
> have now when it comes to saying what is true in mathematics.
Be careful, or you'll have to browse this website:
http://zapatopi.net/afdb.html :-)
>
>> >
>> > Notice that with f=3, the constant term P(0) = u^2(3x + 3u) =
>> > 3u^2(x+u), so it *still* has a factor that is 3, and that's why I
>> > always have the condition that f be coprime to 3.
>> >
>> > Here, however, I'm hoping it'll help to point out why that requirement
>> > is there, and what happens if you ignore it.
>> >
>> > Well then, what are some posters trying to convince you about
>> >
>> > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>> >
>> > 3(-1+mf^2 )x u^2 + u^3 f)?
>> >
>> > They're trying to convince you that there is a mathematical limitation
>> > based on reducibility over Q that determines how f^2 can divide
>> > through when you have the factorization
>> >
>> > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>>
>> I think there's some confusion. P(m)'s 0 term is in fact u^3*f^3
>> when multiplied out.
>
> Nope. Setting m=0, gives P(0) = 3xu^2 + u^3 f = u^2(3x + uf).
I did not say P(0). I said P(m)'s 0 term, which probably needs
to be clarified to P(m)'s x^0 term -- the constant.
If one assumes P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
then the x^0 term is by necessity u^3f^3 -- which turns
out to in fact be the case as you've defined P(m) in a
certain way. The x^3 term is by necessity a_1a_2a_3.
(Since I don't know what the a's are I can't go much
further although I can equate the product thereof to the
term in front of x^3, if I wished to, and work out the
rest of the terms to establish 3 equations in 3 unknowns
relating the a's.)
Apologies if that wasn't clear. As you can see the f^2 factor
waltzes in again; one has to be careful if one drops it from
the intermediate computations. This is fine.
As for your computation of P(0) -- you've simply left out the
f^2 term, from the looks of it. Call that an "oopsie". :-)
>
> What's fascinating about it is that you can see echoes of the
> factorization
>
> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
> as it's clear that at m=0, two and only two of the a's equal 0, as
> that's the only way to get that "u^2" in u^2(3x+uf).
P(m)'s x^3 term is f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3.
P(m)'s x^2 term is always zero.
P(0)'s x^3 term and x^2 terms are therefore both zero.
Therefore exactly two of the a's are 0, as P(0) is
a linear equation in x, not a cubic one.
I wouldn't really expect any other result in that case. :-)
>
> It's actually rather fascinating. Which just makes it that much
> clearer that mathematicians have been avoiding interesting *math*
> instead choosing to focus on obscuring recognition of its validity,
> either directly in posts attacking my me or my work, or indirectly by
> ignoring my work.
>
>> >
>> > and the first question that should come to you is, how could a
>> > constant factor be constrained by reducibility over rationals?
>> >
>> > Now that question is resolvable, but I know the answer is in my favor,
>> > so mathematicians are avoiding even letting you know that IS the
>> > question, and instead those who post work to confuse.
>> >
>> > Now given that I know I have a short proof of Fermat's Last Theorem,
>> > and that mathematicians have been avoiding dealing with reality, while
>> > some posters have gotten away with *deliberately* confusing people,
>> > why would I quit talking about my proof of FLT?
>> >
>> > If you'd found a short proof of Fermat's Last Theorem, would you quit
>> > talking about it?
>>
>> Depends on how many demonstrable errors there were in the proof.
>> My short perusal through your webpages suggests that you might
>> want to clarify your thinking and/or show your work a bit more, as
>> you leap from equation to equation without grinding it out in some
>> cases. I'd have to look to be more specific at this point.
>
> Being specific is important, otherwise your comments can't be put into
> context.
>
> Giving what I've seen I'm not willing to just be trusting.
I like the way Reagan put it: "Trust but verify". :-)
>
> And the great thing about mathematics is that I don't have to be.
True, but one does have to be a little more careful at times.
Euclid made at least one error in some of his proofs, and
apparently the diagonizalization proof of Cantor "proving"
the uncountability of the reals needs shoring up as well.
>
> FYI my website is http://groups.msn.com/AmateurMath
>
> so the proof is out there.
>
>> You might profit by studying Andrew Wiles' proof as well. I don't
>> know if it's on the Web.
>
> Why?
Mostly because he proved Fermat's last theorem. It may come
down to whose proof is simpler, but this sort of thing
occasionally happens in mathematics: two people, working
independently, discover a very similar proof, method,
or identity.
Check out the history of solving x^3 + ax^2 + bx + c,
for example, in
http://mathworld.wolfram.com/CubicEquation.html
(along with the actual solution :-) ).
>
>
> James Harris
--
#191, ewi...@earthlink.net
It's still legal to go .sigless.
The Ghost In The Machine
<3c65f87.03080...@posting.google.com>:
> nora...@hotmail.com (Nora Baron) wrote in message news:<36024859.03080...@posting.google.com>...
>> jst...@msn.com (James Harris) wrote in message news:<3c65f87.03080...@posting.google.com>...
>> > nora...@hotmail.com (Nora Baron) wrote in message news:<36024859.03080...@posting.google.com>...
>
> <deleted>
>
>> > > Not even close. First, in this case, f factors out of your
>> > > polynomial 3 times, not 2 times as in the cases you were
>> > > considering (f <> 3). This is a special case of no interest
>> > > to you or me. It is irrelevant. The Galois argument does not
>> > > apply here. No claims based on that argument are 'destroyed'.
>> > > Second, only one of the proofs that you are wrong in the cases
>> > > where f <> 3 is dependent on Galois Theory. The other proofs are
>> > > based on an elementary theorem from algebraic number theory,
>> > > which you have previously accepted. All of the proofs *do*
>> > > require irreducibility of P(x)/f^2.
>> >
>> > Well, it IS the case that for f=sqrt(2) only *two* of the a's have a
>> > factor that is sqrt(2), so your claim that it is otherwise is false.
>> >
>>
>> In your main applications, as in your "proof" of FLT,
>> f is an integer. I assumed that here.
>
> Your assumption is irrelevant to that fact, as you have tried to
> confuse people by working to convince that m=0 is a special case.
Unfortunately, if one is solving for the a's in
P(m) = (a_1x + uf)(a_2x + uf)(a_3x + uf),
m=0 *is* a special case as exactly two of the a's
become zero. For m != 0, none of the a's are zero.
0 introduces problems in factorization, as you may well appreciate.
[snip for brevity]
> Previous to that I said that your position requires that people
> believe that given
>
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>
> 3(-1+mf^2 )x u^2 + u^3 f) =
>
> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
> that some f^2 divides off as some function of m, or variable dependent
> on m.
I'm leaving this bit in merely to define P(m), for those who may
wander in later. :-)
>
> Splitting that sentence up with your own comments and then posting as
> if I'd said something else is clearly dishonest.
>
>
> James Harris
Bill Dubuque
>
> Lemma. Let R be the ring of all algebraic integers, and let a, b, c be
Readers might enjoy a proof using Gauss' Lemma.
Proof: 1 = (a,c), (b,c)
=> primi a+cx , b+cx primi := primitive
=> primi (a+cx)*(b+cx) by Gauss' Lemma
= ab+c(ax+bx+cxx)
=> 1 = (ab,c) [else product isn't primitive] QED
For an ideal-theoretic proof see my prior post [0]; see also [1].
Methods like above are the heart of Kronecker's theory of forms,
which, in essence, represents ideals as contents of polynomials.
Nowadays this construction is known as the Kronecker function ring,
e.g. see Gilmer's book: Multiplicative Ideal Theory.
Recall that Gauss' Lemma holds true over any Bezout or GCD domain,
indeed it is true in almost all the generalizations of GCD domains
mentioned in my prior post [2]. In fact it is easy to prove that
Gauss' Lemma is equivalent to (a,b) = (a,c) = 1 => (a,bc) = 1
in any integral domain; here (a,b) = 1 means c|a,b => c|1
i.e. coprime, not comaximal. The proof is just a minor variation
of the classic proof of Gauss' Lemma, e.g. see [3, Theorem 3.1].
-Bill Dubuque
[0] http://groups.google.com/groups?selm=y8zissw5xpv.fsf%40nestle.ai.mit.edu
[1] http://groups.google.com/groups?threadm=y8zel3h807z.fsf%40nestle.ai.mit.edu
[2] http://groups.google.com/groups?selm=y8z8yto8k85.fsf%40nestle.ai.mit.edu
[3] Anderson, D. D.(1-IA); Quintero, R. O.(YV-LAND2) MR 98i:13001
Some generalizations of GCD-domains.
Factorization in integral domains (Iowa City, IA, 1996), 189-195,
Lecture Notes in Pure and Appl. Math., 189, Dekker, New York, 1997.
James Harris
> In article <3c65f87.03080...@posting.google.com>,
> James Harris <jst...@msn.com> wrote:
>
> [.snip.]
>
> >> >theory:
> >> >
> >> > THEOREM: If r is a root of a non-monic polynomial
> >> > with integer coefficients, ***irreducible*** over
> >> > the rationals, the r cannot be an algebraic integer.
> >> >
> >> >and conclude that c1 cannot be an algebraic integer.
> >>
> >> You must also include the hypothesis that the polynomial is
> >> primitive. Since nonzero constants are units in Q[x], they are not
> >> considered nontrivial factors, so the hypothesis must be explicitly
> >> included.
> >
> >That is correct. And it is true that c1 cannot be an algebraic
> >integer.
> >
> >That has not been under debate.
>
> Which is why I said I was "nitpicking": pointing out a minor error
> that is well understood.
It is true that c1 cannot be an algebraic integer.
> >It's also NOT under debate as to whether or not given
> >
> > 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)
> >
> >any a's exist, within the ring of algebraic integers, such that
> >sqrt(5) is a factor of them in that ring.
>
> This is rather confused. You have a "not" at the beginning, a "whether
> or not" after that, and a qualifier "any" for the a's. It's pretty
> close to nonsense. What you are really saying, presumably, is:
>
> "Given a1, a2, a3 algebraic integers such that
>
> 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)
>
> [as a polynomial identity], then none of a_1,a_2,a_3 are multiples
> (in the ring of algebraic integers) of sqrt(5)."
It is true that neither a_1, a_2, nor a_3 has sqrt(5) as a factor
***in the ring of algebraic integers***.
> This is also true, and has been established.
Yup as I've stated.
> >The problem is that neither a_1, a_2, nor a_3 have ANY non-unit
> >factors in common with 5 in the ring of algebraic integers.
>
> And that's false. I am pretty sure that Dale produced explicit common
> factors; but in any case, your claim here is certainly false, since
> their product is not coprime to 65.
I've proven it true that neither a_1, a_2 nor a_3 have ANY non-unit
factors in common with 5 in the ring of algebraic integers.
> Lemma. Let R be the ring of all algebraic integers, and let a, b, c be
> any elements of R. If a and b are coprime to c, then a*b is coprime to
> c.
>
> Proof. We use the characterization of coprime valid for commutative
> rings with 1: a and b are coprime in R if and only if there exist x
> and y in R such that ax+by = 1.
By that definition only *one* of the a's is coprime to 5, but none of
them has a factor in common with 5 either.
The ring of algebraic integers is really screwed up.
For those who don't understand, consider that in the ring of evens,
which does not have 1, you can't use that definition of coprime that
Arturo Magidin gives, though it is, interestingly enough, true that in
fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is
not in the ring.
However, rather than use dueling definitions or argue about
definitions I can simply switch to saying that 2 does not share
non-unit factors in the ring of evens with 6.
> Since a and c are coprime by assumption, there exist n and m in R such
> that an+cm = 1. Since b and c are coprime by assumption, there exist r
> and s in R such that br+cs = 1.
>
> Multiplying both together, we have
>
> 1 = (an+cm)(br+cs)
> = abrn + acns + cbmr + c^2*ms
> = ab(rn) + c(ans + bmr + cms).
>
> Let x = rn, y = ans+bmr+cms. Then x and y are algebraic integrs, and
> ab*x + c*y = 1. Therefore, ab and y are coprime. QED
>
> So, assume you were correct and neither a_1, a_2, nor a_3 have ANY
> non-unit factors in common with 5 in the ring of algebraic
> integers. Then, by the lemma, neither does a1*a_2; and applying the
> lemma again, neither does a_1*a_2*a_3. But a_1*a_1*a_3 = 65, which
> clearly has 5 as a nonunit common factor with 5. This contradicts the
> assumption that none of a_1, a_2, a_3 have common non-unit factors
> with 5 in the ring of algebraic integers. Therefore, your assertion is
> false.
Well by your definition of "coprime" NONE of the a's have a factor in
common with 5, in the ring of algebraic integers, and you cannot prove
that any of them do.
Now if you don't want to call that "coprime" fine. It doesn't change
the situation.
What I can do is show that with a very quick argument using basic
algebra as I've done.
Consider, in the ring of algebraic integers,
P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f).
Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization
P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
where w_1 w_2 w_3 = f, and
b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),
and at m=0
P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
so two of the b's must equal 0, which means
P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)
which is
P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)
proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves
b_3 = 3.
Essentially objections to how f^2 divides off now come down to
claiming that the w's are functions of m, but consider that w_1 w_2 =
1, when m=0, if f is coprime to 3.
But that was an arbitrary choice, so let f=3.
Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m.
That is, the w's are now all constant with regard to m and have the
same value no matter what the value of m is.
Therefore, the factorization is
P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f =
(b_1 x + u)(b_2 x + u)(b_3 x + uf)
where you'll notice that the b's are algebraic integers with m=1,
f=sqrt(2), but that's a special case as generally they are not, and
using m=1, f=sqrt(5), gives
65x^3 - 12xy^2 + y^3
which results in b's that are NOT algebraic integers.
I've found the Ring of Objects which includes the ring of algebraic
integers, and does not have this problem, as the b's are all included
in it.
The Ring of Objects is the set of all numbers where -1 and 1 are the
only members that are both a unit, i.e. factor of 1, and an integer,
where no non-unit member is a factor of any two integers that are
coprime.
That definition and more is linked to from my primary website
http://groups.msn.com/AmateurMath
where you can also find information on my other math research.
James Harris
C. Bond
[snip]
Earth to James...
Please post one, count 'em: *one*, number which belongs in the ring of algebraic integers but which
somehow got 'left out'. You have repeatedly claimed that the ring of algebraic integers is incomplete
because it fails to include numbers which belong in it. With all due contempt I believe you have some
obligation to produce at least one of them -- a number, mind you, not an ambiguous expression consisting
of ill-defined symbols.
Name one such number, if you can. If not, stop claiming that such numbers exist. (In your breathlessly
anticipated reply, please do not include terminology like: "amazing", "fascinating", and "odd". These
expose your mental state, not the underlying mathematics of the subject at hand.)
--
There are two things you must never attempt to prove: the unprovable -- and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
Arturo Magidin
James Harris <jst...@msn.com> wrote:
>mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<bh3fit$1l1l$1...@agate.berkeley.edu>...
>> In article <3c65f87.03080...@posting.google.com>,
>> James Harris <jst...@msn.com> wrote:
>>
>> [.snip.]
>>
>> >> >theory:
>> >> >
>> >> > THEOREM: If r is a root of a non-monic polynomial
>> >> > with integer coefficients, ***irreducible*** over
>> >> > the rationals, the r cannot be an algebraic integer.
>> >> >
>> >> >and conclude that c1 cannot be an algebraic integer.
>> >>
>> >> You must also include the hypothesis that the polynomial is
>> >> primitive. Since nonzero constants are units in Q[x], they are not
>> >> considered nontrivial factors, so the hypothesis must be explicitly
>> >> included.
>> >
>> >That is correct. And it is true that c1 cannot be an algebraic
>> >integer.
>> >
>> >That has not been under debate.
>>
>> Which is why I said I was "nitpicking": pointing out a minor error
>> that is well understood.
>
>It is true that c1 cannot be an algebraic integer.
Which was not what I addressed. I included the stuff before it to
provide context, and then pointed out that a hypothesis was missing
from the Theorem quoted.
What exactly is your problem with that? Are you asserting that the
Theorem was correct as written? Are you asserting that the theorem is
wrong as corrected according to the mistake I pointed out?
Or are you simply committing the fallacy of Argumentum ad logicam?
http://www.infidels.org/news/atheism/logic.html#logicam
[.snip.]
>> > 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)
>> >
>> >any a's exist, within the ring of algebraic integers, such that
>> >sqrt(5) is a factor of them in that ring.
>>
>> This is rather confused. You have a "not" at the beginning, a "whether
>> or not" after that, and a qualifier "any" for the a's. It's pretty
>> close to nonsense. What you are really saying, presumably, is:
>>
>> "Given a1, a2, a3 algebraic integers such that
>>
>> 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)
>>
>> [as a polynomial identity], then none of a_1,a_2,a_3 are multiples
>> (in the ring of algebraic integers) of sqrt(5)."
>
>It is true that neither a_1, a_2, nor a_3 has sqrt(5) as a factor
>***in the ring of algebraic integers***.
>
>> This is also true, and has been established.
>
>Yup as I've stated.
As I've ->PROVEN<- (and was proven by Bengt, and Dale, and Dik, and
Nora, and David, and many others), and as you DENIED, mightily, for
months on end. But thanks for finally coming around, even if now you
are trying to claim credit for a result you do not understand and
cannot prove.
>> >The problem is that neither a_1, a_2, nor a_3 have ANY non-unit
>> >factors in common with 5 in the ring of algebraic integers.
>>
>> And that's false. I am pretty sure that Dale produced explicit common
>> factors; but in any case, your claim here is certainly false, since
>> their product is not coprime to 65.
>
>I've proven it true that neither a_1, a_2 nor a_3 have ANY non-unit
>factors in common with 5 in the ring of algebraic integers.
This is paraphrased from the post
http://groups.google.com/groups?selm=3F148963.6000800%40farir.com
by W. Dale Hall, made on July 15, 2003.
Take the polynomial 65x^3 - 12x + 1, and factor it as
65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1)
with a1, a2, a3 (necessarily) algebraic integers (in fact, minus the
roots of x^3 - 12x^2 + 65). Let z be any root of that polynomial. That
is, z will be either -a1, -a2, or -a3. It is trivial that any common
factor between z and 5 will be a common factor of the corresponding ai.
Take the following three polynomials
q(x) = 8 x^2 - 76 x - 185
r(x) = 8 x^2 - 4 x - 45
s(x) = 4 x^2 - 37 x - 104
Then any product of q(z), r(z), s(z) and integers will be an algebraic
integer, necessarily.
We have that
q(z)*r(z) = 64 z^4 - 640 z^3 - 1536 z^2 + 4160 z + 8325
Note that
(64 x + 128)*(x^3 - 12 x^2 + 65)
= 64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8320
so (65z + 128)(z^3 - 12z^2 + 65) = 64z^4 - 650z^3 - 1536z^2 +
4160z+8320.
But z^3-12z^2+65 = 0. So 64z^4 - 650z^3 - 1536z^2 +4160z+8320 = 0.
Therefore,
q(z)*r(z) = 64 z^4 - 640 z^3 - 1536 z^2 + 4160 z + 8325
= (64 z^4 - 640 z^3 - 1536 z^2 + 4160 z + 8320) + 5
= 5.
Likewise, take
r(z)*s(z) = 32 z^4 - 312 z^3 - 864 z^2 + 2081 z + 4680
This time, note that
(32 x + 72)*(x^3 - 12 x^2 + 65)
= 32 x^4 - 312 x^3 - 864 x^2 + 2080 x + 4680
So (32z+72)*(z^3-12z^2+65)
= 32z^4 - 312z^3 - 864z^2 + 2080z + 4680.
But since z^3-12z^2+65 is equal to 0, it follows that
32z^4 - 312z^3 - 864z^2 + 2080z + 4680=0. Therefore,
r(z)*s(z) = 32z^4 - 312z^3 - 864z^2 + 2081z + 4680
= (32z^4 - 312z^3 - 864z^2 + 2080z + 4680) + z
= z.
Therefore, r(z), which is an algebraic integer, is a common factor of
5 and of z. It only remains to show that it is not a unit.
The claim is that r(z) is a root of f(x) = x^3 - 969 x^2 + 315 x +5.
If this is so, then r(z) is not a unit in the ring of all algebraic
integers, since this is a monic irreducible polynomial with integer
coefficients whose constant term is neither 1 nor -1.
As Dale noted, we have
f(r(z)) = (r(z))^3 - 969 (r(z))^2 + 315 (r(z)) + 5
= (8 z^2 - 4 z - 45)^3 - 969 (8 z^2 - 4 z - 45)^2
+ 315 (8 z^2 - 4 z - 45) + 5
= 512 z^6 - 768 z^5 - 70272 z^4 + 70592 z^3
+ 731136 z^2 - 374400 z - 2067520.
Letting w(z) = 512 z^3 + 5376 z^2 - 5760 z - 31808
we have that
p(z)*w(z) = 512 z^6 - 768 z^5 - 70272 z^4 + 70592 z^3
+ 731136 z^2 - 374400 z - 2067520
and therefore, f(r(z)) = p(z)*w(z). But we know that p(z)=0, so
f(r(z))=0. This proves that r(z) is not a unit, and yet is a common
factor of z and 5.
Letting z be -a1, -a2, and -a3, in turn, you obtain common factors of
EACH of a1, a2, a3 with 5 in the ring of all algebraic integers, which
are not units.
>> Lemma. Let R be the ring of all algebraic integers, and let a, b, c be
>> any elements of R. If a and b are coprime to c, then a*b is coprime to
>> c.
>>
>> Proof. We use the characterization of coprime valid for commutative
>> rings with 1: a and b are coprime in R if and only if there exist x
>> and y in R such that ax+by = 1.
>
>By that definition
That property is EQUIVALENT to being coprime in the algebraic
integers, which is EQUIVALENT to having no common factors other than units.
> only *one* of the a's is coprime to 5, but none of
>them has a factor in common with 5 either.
Nonsense. By that definition, NONE of the a's are coprime to 5, and
ALL have common factors with 5 which are not units. See the explicit
calculations above.
>The ring of algebraic integers is really screwed up.
Something is, but it is not the ring of algebraic integers.
>For those who don't understand, consider that in the ring of evens,
>which does not have 1, you can't use that definition of coprime that
>Arturo Magidin gives, though it is, interestingly enough, true that in
>fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is
>not in the ring.
The ring 2Z is not a ring with unit, so the characterization given
above does not apply. As it happens, 2 and 6 are coprime in the ring
in question, but NOT for the incoherent reason you give. They are
coprime because the ideal generated by 2 is maximal, contains the
ideal generated by 6, and is not prime.
>However, rather than use dueling definitions or argue about
>definitions I can simply switch to saying that 2 does not share
>non-unit factors in the ring of evens with 6.
But it does not matter: above I have reproduced EXPLICIT common
factors of each of a1, a2, and a3 with 5 in the ring of algebraic
integers, and none of them are units. Your claim is as dead as it
always was.
[.snip.]
>> So, assume you were correct and neither a_1, a_2, nor a_3 have ANY
>> non-unit factors in common with 5 in the ring of algebraic
>> integers. Then, by the lemma, neither does a1*a_2; and applying the
>> lemma again, neither does a_1*a_2*a_3. But a_1*a_1*a_3 = 65, which
>> clearly has 5 as a nonunit common factor with 5. This contradicts the
>> assumption that none of a_1, a_2, a_3 have common non-unit factors
>> with 5 in the ring of algebraic integers. Therefore, your assertion is
>> false.
>
>Well by your definition of "coprime"
It's not "my definition." It's a property which is equivalent to the
STANDARD definition (which is NOT "has no common factors other than
units", as you have been told numerous times), and which in turn is
equivalent, IN THE RING OF ALL ALGEBRAIC INTEGERS, to the property
that they have no common factors.
> NONE of the a's have a factor in
>common with 5, in the ring of algebraic integers, and you cannot prove
>that any of them do.
Weird. Because above I exhibited explicit factors.
>Now if you don't want to call that "coprime" fine. It doesn't change
>the situation.
No, it does not. You're wrong.
[.snip.]
You continue to argue that a singular case implies a general case, and
to argue that a numeric identity implies a polynomial one. Both are
colossal errors.
Arturo Magidin
Bill Dubuque <w...@nestle.ai.mit.edu> wrote:
>Arturo Magidin <mag...@math.berkeley.edu> wrote:
>>
>> Take your favorite quadratic extension of Q. Let p be a rational prime
>> which splits in the ring of integers of K, (p)=PQ, with P<>Q. One can
>> then invert elements which are in P but not in Q to obtain a ring
>> larger than the ring of integers of K, which still satisfies the
>> condition that the only rational integers which are units are 1 and -1.
>
>That's the point of the example I posted.
Yes; I believe I mentioned that the example was suggested by the post
quoted (yours); at least the previous version, which contained my
incorrect followup to the example, did. Sorry if I wasn't clearer in
giving credit where it is due.
Will Twentyman
> mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<bh3fit$1l1l$1...@agate.berkeley.edu>...
>
>>In article <3c65f87.03080...@posting.google.com>,
>>James Harris <jst...@msn.com> wrote:
>>
>>>The problem is that neither a_1, a_2, nor a_3 have ANY non-unit
>>>factors in common with 5 in the ring of algebraic integers.
>>
>>And that's false. I am pretty sure that Dale produced explicit common
>>factors; but in any case, your claim here is certainly false, since
>>their product is not coprime to 65.
>
>
> I've proven it true that neither a_1, a_2 nor a_3 have ANY non-unit
> factors in common with 5 in the ring of algebraic integers.
>
And he proves (below) that they do. At this point you have to do more
than point at your own proof. You must find the error in his.
>
>>Lemma. Let R be the ring of all algebraic integers, and let a, b, c be
>>any elements of R. If a and b are coprime to c, then a*b is coprime to
>>c.
>>
>>Proof. We use the characterization of coprime valid for commutative
>>rings with 1: a and b are coprime in R if and only if there exist x
>>and y in R such that ax+by = 1.
>
>
> By that definition only *one* of the a's is coprime to 5, but none of
> them has a factor in common with 5 either.
Do you even understand what he's saying? The application would be: If
a_1 and a_2 are coprime to 5, then so is a_1*a_2. Extending it, if a_1,
a_2, and a_3 are coprime to 5, then a_1*a_2 and a_3 are coprime to 5,
which means a_1*a_2*a_3 is coprime to 5.
>
> The ring of algebraic integers is really screwed up.
No, your understanding of them is.
>
> For those who don't understand, consider that in the ring of evens,
> which does not have 1, you can't use that definition of coprime that
> Arturo Magidin gives, though it is, interestingly enough, true that in
> fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is
> not in the ring.
Why do you make unnecessary side trips to bizarre rings?
>
> However, rather than use dueling definitions or argue about
> definitions I can simply switch to saying that 2 does not share
> non-unit factors in the ring of evens with 6.
>
>
>>Since a and c are coprime by assumption, there exist n and m in R such
>>that an+cm = 1. Since b and c are coprime by assumption, there exist r
>>and s in R such that br+cs = 1.
>>
>>Multiplying both together, we have
>>
>>1 = (an+cm)(br+cs)
>> = abrn + acns + cbmr + c^2*ms
>> = ab(rn) + c(ans + bmr + cms).
>>
>>Let x = rn, y = ans+bmr+cms. Then x and y are algebraic integrs, and
>>ab*x + c*y = 1. Therefore, ab and y are coprime. QED
>>
>>So, assume you were correct and neither a_1, a_2, nor a_3 have ANY
>>non-unit factors in common with 5 in the ring of algebraic
>>integers. Then, by the lemma, neither does a1*a_2; and applying the
>>lemma again, neither does a_1*a_2*a_3. But a_1*a_1*a_3 = 65, which
>>clearly has 5 as a nonunit common factor with 5. This contradicts the
>>assumption that none of a_1, a_2, a_3 have common non-unit factors
>>with 5 in the ring of algebraic integers. Therefore, your assertion is
>>false.
>
>
> Well by your definition of "coprime" NONE of the a's have a factor in
> common with 5, in the ring of algebraic integers, and you cannot prove
> that any of them do.
No, what he has said is that if a_1, a_2, a_3 are coprime to 5, then
their product is as well, which means a_1*a_2*a_3 *cannot* equal 65.
This shows that there is a mistake in your work since you are claiming
the product of three algebraic integers coprime to 5 *is* 65.
>
> Now if you don't want to call that "coprime" fine. It doesn't change
> the situation.
>
> What I can do is show that with a very quick argument using basic
> algebra as I've done.
You keep posting the same text as if it hasn't been riddled with holes.
How about dealing with people's counter-arguments directly instead of
posting the same thing over and over?
[usual "proof" deleted]
>
> I've found the Ring of Objects which includes the ring of algebraic
> integers, and does not have this problem, as the b's are all included
> in it.
>
> The Ring of Objects is the set of all numbers where -1 and 1 are the
> only members that are both a unit, i.e. factor of 1, and an integer,
> where no non-unit member is a factor of any two integers that are
> coprime.
How does the "Ring of Objects" differ from the algebraic integers?
Are you sure your definition is consistent?
In which ring are you referring to the two integers being coprime?
--
Will Twentyman
email: wtwentyman at copper dot net
James Harris
> James Harris wrote:
> > mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<bh3fit$1l1l$1...@agate.berkeley.edu>...
> >
> >>In article <3c65f87.03080...@posting.google.com>,
> >>James Harris <jst...@msn.com> wrote:
> >>
>
> >>>The problem is that neither a_1, a_2, nor a_3 have ANY non-unit
> >>>factors in common with 5 in the ring of algebraic integers.
> >>
> >>And that's false. I am pretty sure that Dale produced explicit common
> >>factors; but in any case, your claim here is certainly false, since
> >>their product is not coprime to 65.
> >
> >
> > I've proven it true that neither a_1, a_2 nor a_3 have ANY non-unit
> > factors in common with 5 in the ring of algebraic integers.
> >
>
> And he proves (below) that they do. At this point you have to do more
> than point at your own proof. You must find the error in his.
He uses a special definition of coprime, and assumes that at least two
of the three numbers is coprime.
However, by his definition of coprime, only one of the numbers is
coprime.
> >
> >>Lemma. Let R be the ring of all algebraic integers, and let a, b, c be
> >>any elements of R. If a and b are coprime to c, then a*b is coprime to
> >>c.
> >>
> >>Proof. We use the characterization of coprime valid for commutative
> >>rings with 1: a and b are coprime in R if and only if there exist x
> >>and y in R such that ax+by = 1.
> >
> >
> > By that definition only *one* of the a's is coprime to 5, but none of
> > them has a factor in common with 5 either.
>
> Do you even understand what he's saying? The application would be: If
> a_1 and a_2 are coprime to 5, then so is a_1*a_2. Extending it, if a_1,
> a_2, and a_3 are coprime to 5, then a_1*a_2 and a_3 are coprime to 5,
> which means a_1*a_2*a_3 is coprime to 5.
By the definition he gave only ONE of the a's is coprime to 5.
The others don't have non unit factors in common with 5, in the ring
of algebraic integers, but each have a factor that is sqrt(5) in a
higher ring.
> >
> > The ring of algebraic integers is really screwed up.
>
> No, your understanding of them is.
Nope. The ring of algebraic integers is really screwed up.
> >
> > For those who don't understand, consider that in the ring of evens,
> > which does not have 1, you can't use that definition of coprime that
> > Arturo Magidin gives, though it is, interestingly enough, true that in
> > fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is
> > not in the ring.
>
> Why do you make unnecessary side trips to bizarre rings?
Arturo Magidin uses a special definition of coprime which depends on
certain assumptions which fail with the ring of algebraic integers,
and I highlight for readers how that definition can also fail in
another ring.
Only one of them is coprime to 5, and the other two should have some
non unit factor in common with 5 but don't because the ring of
algebraic integers is flawed.
> >
> > Now if you don't want to call that "coprime" fine. It doesn't change
> > the situation.
> >
> > What I can do is show that with a very quick argument using basic
> > algebra as I've done.
>
> You keep posting the same text as if it hasn't been riddled with holes.
> How about dealing with people's counter-arguments directly instead of
> posting the same thing over and over?
>
> [usual "proof" deleted]
Deleting out a mathematical argument does not make it go away.
> >
> > I've found the Ring of Objects which includes the ring of algebraic
> > integers, and does not have this problem, as the b's are all included
> > in it.
> >
> > The Ring of Objects is the set of all numbers where -1 and 1 are the
> > only members that are both a unit, i.e. factor of 1, and an integer,
> > where no non-unit member is a factor of any two integers that are
> > coprime.
>
> How does the "Ring of Objects" differ from the algebraic integers?
>
> Are you sure your definition is consistent?
>
> In which ring are you referring to the two integers being coprime?
The Ring of Objects is a higher ring than the ring of algebraic
integers, which doesn't have its problems.
The two integers are coprime in the Ring of Objects.
The definition removes the possibility of the fundamental
contradiction which distinguishes a field from a ring, which is that
in a field coprimeness does not exist, while in the ring of integers,
it does.
So fields are at odds with the ring of integers.
I've simply highlighted the crucial difference between them.
James Harris
Norman Stern
This post is classic Harris, marred only by the incoherent mumblings
of others. I remove those mumblings to reveal James' undiluted efflux:
#######
> He uses a special definition of coprime, and assumes that at least two
> of the three numbers is coprime.
>
> However, by his definition of coprime, only one of the numbers is
> coprime.
#######
> By the definition he gave only ONE of the a's is coprime to 5.
>
> The others don't have non unit factors in common with 5, in the ring
> of algebraic integers, but each have a factor that is sqrt(5) in a
> higher ring.
#######
> Nope. The ring of algebraic integers is really screwed up.
#######
> Arturo Magidin uses a special definition of coprime which depends on
> certain assumptions which fail with the ring of algebraic integers,
> and I highlight for readers how that definition can also fail in
> another ring.
#######
> Only one of them is coprime to 5, and the other two should have some
> non unit factor in common with 5 but don't because the ring of
> algebraic integers is flawed.
#######
> Deleting out a mathematical argument does not make it go away.
#######
> The Ring of Objects is a higher ring than the ring of algebraic
> integers, which doesn't have its problems.
>
> The two integers are coprime in the Ring of Objects.
>
> The definition removes the possibility of the fundamental
> contradiction which distinguishes a field from a ring, which is that
> in a field coprimeness does not exist, while in the ring of integers,
> it does.
>
> So fields are at odds with the ring of integers.
>
> I've simply highlighted the crucial difference between them.
>
> James Harris
#######
Arturo Magidin
James Harris <jst...@msn.com> wrote:
>Will Twentyman <wtwen...@read.my.sig> wrote in message news:<3f369f7b$1_2@newsfeed>...
>> James Harris wrote:
>> > mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<bh3fit$1l1l$1...@agate.berkeley.edu>...
>> >
>> >>In article <3c65f87.03080...@posting.google.com>,
>> >>James Harris <jst...@msn.com> wrote:
>> >>
>>
>> >>>The problem is that neither a_1, a_2, nor a_3 have ANY non-unit
>> >>>factors in common with 5 in the ring of algebraic integers.
>> >>
>> >>And that's false. I am pretty sure that Dale produced explicit common
>> >>factors; but in any case, your claim here is certainly false, since
>> >>their product is not coprime to 65.
>> >
>> >
>> > I've proven it true that neither a_1, a_2 nor a_3 have ANY non-unit
>> > factors in common with 5 in the ring of algebraic integers.
>> >
>>
>> And he proves (below) that they do. At this point you have to do more
>> than point at your own proof. You must find the error in his.
>
>He uses a special definition of coprime, and assumes that at least two
>of the three numbers is coprime.
No, I use the standard definition of coprime: two elements a and b are
coprime in R if and only if the ideals (a) and (b) are coprime, if and
only if there is no prime ideal of R which contains both (a) and (b).
It is a THEOREM that in a commutative ring with 1, two elements a and
b are coprime if and only if there exist x and y in R such that
ax+by=1.
What is the definition YOU are using?
The definition you are using is "a and b are coprime if and only if
they have no non-unit common factors."
In the ring of all algebraic integers, the two are
EQUIVALENT. However, your definition is WEAKER than the correct
meaning of the term: in Z[sqrt(-5)], 1+sqrt(-5) and 2 are coprime
under your incorrect usage, but they are not coprime under correct usage.
>However, by his definition of coprime, only one of the numbers is
>coprime.
"Coprime" to ->what<-?
And prove it.
[.snip.]
>> >>Lemma. Let R be the ring of all algebraic integers, and let a, b, c be
>> >>any elements of R. If a and b are coprime to c, then a*b is coprime to
>> >>c.
>> >>
>> >>Proof. We use the characterization of coprime valid for commutative
>> >>rings with 1: a and b are coprime in R if and only if there exist x
>> >>and y in R such that ax+by = 1.
>> >
>> >
>> > By that definition only *one* of the a's is coprime to 5, but none of
>> > them has a factor in common with 5 either.
>>
>> Do you even understand what he's saying? The application would be: If
>> a_1 and a_2 are coprime to 5, then so is a_1*a_2. Extending it, if a_1,
>> a_2, and a_3 are coprime to 5, then a_1*a_2 and a_3 are coprime to 5,
>> which means a_1*a_2*a_3 is coprime to 5.
>
>By the definition he gave only ONE of the a's is coprime to 5.
Prove it.
>The others don't have non unit factors in common with 5, in the ring
>of algebraic integers, but each have a factor that is sqrt(5) in a
>higher ring.
Which doesn't prove anything. If the others "don't have non-unit
factors in common with 5" in the ring of all algebraic integers, then
they would be coprime under the correct definition as well.
[.snip.]
>> > For those who don't understand, consider that in the ring of evens,
>> > which does not have 1, you can't use that definition of coprime that
>> > Arturo Magidin gives, though it is, interestingly enough, true that in
>> > fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is
>> > not in the ring.
>>
>> Why do you make unnecessary side trips to bizarre rings?
>
>Arturo Magidin uses a special definition of coprime which depends on
>certain assumptions which fail with the ring of algebraic integers,
I use the STANDARD definition; it is you who uses a "SPECIAL"
one.
Please list, explicitly:
1. The "special definition";
2. The one you use;
3. The "certain assumptions" on which my supposedly special definition
depends.
4. A proof that they fail in the ring of algebraic integers.
It is not enough to say "the ring is really screwed up", because that
is supposed to be your conclusion, not your hypothesis.
[.snip.]
Will Twentyman
> Will Twentyman <wtwen...@read.my.sig> wrote in message news:<3f369f7b$1_2@newsfeed>...
>
>>James Harris wrote:
>>
>>>mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<bh3fit$1l1l$1...@agate.berkeley.edu>...
>>>
>>>>In article <3c65f87.03080...@posting.google.com>,
>>>>James Harris <jst...@msn.com> wrote:
>>
>>>>>The problem is that neither a_1, a_2, nor a_3 have ANY non-unit
>>>>>factors in common with 5 in the ring of algebraic integers.
>>>>
>>>>And that's false. I am pretty sure that Dale produced explicit common
>>>>factors; but in any case, your claim here is certainly false, since
>>>>their product is not coprime to 65.
>>>
>>>I've proven it true that neither a_1, a_2 nor a_3 have ANY non-unit
>>>factors in common with 5 in the ring of algebraic integers.
>>
>>And he proves (below) that they do. At this point you have to do more
>>than point at your own proof. You must find the error in his.
>
> He uses a special definition of coprime, and assumes that at least two
> of the three numbers is coprime.
What definition of coprime would you suggest that he use? What
definition of coprime are you using?
> However, by his definition of coprime, only one of the numbers is
> coprime.
I'll address this below.
>>>>Lemma. Let R be the ring of all algebraic integers, and let a, b, c be
>>>>any elements of R. If a and b are coprime to c, then a*b is coprime to
>>>>c.
>>>>
>>>>Proof. We use the characterization of coprime valid for commutative
>>>>rings with 1: a and b are coprime in R if and only if there exist x
>>>>and y in R such that ax+by = 1.
>>>
>>>
>>>By that definition only *one* of the a's is coprime to 5, but none of
>>>them has a factor in common with 5 either.
>>
>>Do you even understand what he's saying? The application would be: If
>>a_1 and a_2 are coprime to 5, then so is a_1*a_2. Extending it, if a_1,
>>a_2, and a_3 are coprime to 5, then a_1*a_2 and a_3 are coprime to 5,
>>which means a_1*a_2*a_3 is coprime to 5.
>
> By the definition he gave only ONE of the a's is coprime to 5.
>
> The others don't have non unit factors in common with 5, in the ring
> of algebraic integers, but each have a factor that is sqrt(5) in a
> higher ring.
Which ring is that? If you mean your "Ring of Objects", please explain
why sqrt(5) is a factor there and not in the algebraic integers.
>>>For those who don't understand, consider that in the ring of evens,
>>>which does not have 1, you can't use that definition of coprime that
>>>Arturo Magidin gives, though it is, interestingly enough, true that in
>>>fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is
>>>not in the ring.
>>
>>Why do you make unnecessary side trips to bizarre rings?
>
> Arturo Magidin uses a special definition of coprime which depends on
> certain assumptions which fail with the ring of algebraic integers,
> and I highlight for readers how that definition can also fail in
> another ring.
What are the assumptions that fail with the ring of algebraic integers?
What definition *should* he be using if this one is not appropriate?
The question becomes: what do you mean by coprime, and why does your
definition differ from his in the algebraic integers?
James, "should be don't" is the same as "don't". What you are saying
could be rephrased as "They are all coprime to 5, but I want 2 of them
not to be." If I'm not understanding what you are saying here, please
explain it. Note: I'm assuming that no non-unit factor in common with 5
is equivalent to coprime to 5 in the algebraic integers. If you
disagree with that, please state why.
>>>Now if you don't want to call that "coprime" fine. It doesn't change
>>>the situation.
>>>
>>>What I can do is show that with a very quick argument using basic
>>>algebra as I've done.
>>
>>You keep posting the same text as if it hasn't been riddled with holes.
>> How about dealing with people's counter-arguments directly instead of
>>posting the same thing over and over?
>>
>>[usual "proof" deleted]
>
> Deleting out a mathematical argument does not make it go away.
No, but you haven't said anything new in a while, and it has nothing to
do with what I was talking about.
>>>I've found the Ring of Objects which includes the ring of algebraic
>>>integers, and does not have this problem, as the b's are all included
>>>in it.
>>>
>>>The Ring of Objects is the set of all numbers where -1 and 1 are the
>>>only members that are both a unit, i.e. factor of 1, and an integer,
>>>where no non-unit member is a factor of any two integers that are
>>>coprime.
>>
>>How does the "Ring of Objects" differ from the algebraic integers?
>>
>>Are you sure your definition is consistent?
>>
>>In which ring are you referring to the two integers being coprime?
>
>
> The Ring of Objects is a higher ring than the ring of algebraic
> integers, which doesn't have its problems.
Aside from the algebraic integers not doing what you think they "should"
do, what problems do they have?
>
> The two integers are coprime in the Ring of Objects.
Ok, so depending on what is in the Ring of Objects, 2 and 3 may or may
not be coprime... how would I check this?
> The definition removes the possibility of the fundamental
> contradiction which distinguishes a field from a ring, which is that
> in a field coprimeness does not exist, while in the ring of integers,
> it does.
Coprimeness exists in a ring, it's just not very interesting.
> So fields are at odds with the ring of integers.
Considering the ring of integers is *not* a field, this should not be
surprising.
> I've simply highlighted the crucial difference between them.
You mean the whole point of this months of discussion has been about the
fact that the ring of integers is not a field? Am I missing something?
> James Harris
Arturo Magidin
Will Twentyman <wtwen...@read.my.sig> wrote:
>James Harris wrote:
>> Will Twentyman <wtwen...@read.my.sig> wrote in message news:<3f369f7b$1_2@newsfeed>...
>>
>>>James Harris wrote:
>>>
>>>>mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<bh3fit$1l1l$1...@agate.berkeley.edu>...
>>>>
>>>>>In article <3c65f87.03080...@posting.google.com>,
>>>>>James Harris <jst...@msn.com> wrote:
>>>
>>>>>>The problem is that neither a_1, a_2, nor a_3 have ANY non-unit
>>>>>>factors in common with 5 in the ring of algebraic integers.
>>>>>
>>>>>And that's false. I am pretty sure that Dale produced explicit common
>>>>>factors; but in any case, your claim here is certainly false, since
>>>>>their product is not coprime to 65.
>>>>
>>>>I've proven it true that neither a_1, a_2 nor a_3 have ANY non-unit
>>>>factors in common with 5 in the ring of algebraic integers.
>>>
>>>And he proves (below) that they do. At this point you have to do more
>>>than point at your own proof. You must find the error in his.
>>
>> He uses a special definition of coprime, and assumes that at least two
>> of the three numbers is coprime.
>
>What definition of coprime would you suggest that he use? What
>definition of coprime are you using?
I believe that James incorrectly believes that "a is coprime to b"
means "a has no nonunit common factors with b." That definition is
equivalent to the standard one in many rings (e.g., in the ring of
integers, in the ring of all algebraic integers, in any PID), but not
in others.
The interesting point, of course, is that the definition James is
using, if it were not equivalent to the usual one, would be useless
for his purposes. James needs the coprime property in order to remove
factors in a congruence, but that step is not valid if you only have
"have no nonunit common factors". You need coprime in the usual sense,
with its equivalent statement that there are x and y such that
ax+by=1. (If the two are equivalent in the ring in question, of
course, there is no problem; but they are not always equivalent).
For instance, in Z[sqrt(-5)], the two statements are not equivalent. 2
and 1+sqrt(-5) have no common factors other than units, but they are
not coprime; we have that (1+sqrt(-5))*(1-sqrt(-5)) = 0 (mod 2) in
Z[sqrt(-5)], but we cannot conclude from the fact that 2 and
1+sqrt(-5) have no factors in common that 1-sqrt(-5)=0 (mod 2); it
clearly does not (all multiples of 2 are of the form 2a+2b*sqrt(-5), a
and b integers).
Rather, the usual argument is that if a*b=0 (mod c) and a and c are
coprime, then we know that c|ab. Since there exist elements x,y such
that ax+cy = 1, multiplying by b we have abx + cby = b, and since c
divides abx and cby, it divides b; thus b=0 (mod c) is a valid
conclusion.
If you want another example that James likes, as he points out
(his conclusion is correct, although his reasoning is incorrect), 2
and 6 are coprime in 2Z: the ideal (2) consists of all integers
divisible by 4, and is maximal; (6) consists of all integers
divisible by 12; so (6) is contained in (2). The only ideal that
contains both (2) and (6) is (2); and (2) is not a prime ideal: 2*2
lies in (2), but 2 does not. Thus, there is no prime ideal of 2Z
containing both (2) and (6), so 2 and 6 are coprime (they are not,
however, comaximal; the two conditions are equivalent in a ring with
1, assuming the Axiom of Choice).
However, even though 2 and 6 are coprime, we cannot cancel in
congruence in this ring. 12=0 (mod 2), since 12=2*6. However, if we
write 12=2*6=0 (mod 2) (in 2Z) and argue that 6 is coprime to 2, and
therefore we can cancel from the congruence, we would get that 2=0
(mod 2) (in 2Z), and that statement is false: there is no element k in
2Z such that 2=0+2k.
So, in the end, the definition which James is using is not the one he
should be using for his purposes, unless he can prove that in the ring
he is working on it is equivalent to the usual one. Never mind that
there is no "ring of objects" as he has defined it.
[.snip.]
>Which ring is that? If you mean your "Ring of Objects", please explain
>why sqrt(5) is a factor there and not in the algebraic integers.
There is no "Ring of Objects" according to his definition. Even
allowing for the possibility that "number" means "complex number" or
"algebraic number", there is no unique ring determined by the
conditions he gives, yet he assumes there is one and only one such ring.
James Harris
<deleted>
> >He uses a special definition of coprime, and assumes that at least two
> >of the three numbers is coprime.
>
> No, I use the standard definition of coprime: two elements a and b are
> coprime in R if and only if the ideals (a) and (b) are coprime, if and
> only if there is no prime ideal of R which contains both (a) and (b).
>
> It is a THEOREM that in a commutative ring with 1, two elements a and
> b are coprime if and only if there exist x and y in R such that
> ax+by=1.
That's fascinating. Why don't you give the theorem, as it must be in
error for the ring of algebraic integers.
And yes, I would like to see that theorem as I'd be fascinated to see
some evidence of how deep the weird algebraic integer ring problem has
gone in terms of what mathematicians believe to be true.
Still my guess is that it depends on assertions about units, where you
base your belief, once again, on assuming that a number that is not a
unit in the ring of algebraic integers must be a non-unit factor in
the ring of algebraic integers.
The problem is fascinating indeed, and easily leads to confusion, if
you aren't careful to be *very* careful with your definitions and
assumptions.
That is, you need to follow the math, not what you *think* is the
truth.
> What is the definition YOU are using?
I've given the following definition for coprime in the proper ring,
which is the ring of objects:
Two objects are coprime if they don't share a non-unit factor e.g. 2
is coprime to 3, while 2 and 4 share 2 as a non-unit factor.
> The definition you are using is "a and b are coprime if and only if
> they have no non-unit common factors."
Yeah, that's basically it.
> In the ring of all algebraic integers, the two are
> EQUIVALENT. However, your definition is WEAKER than the correct
> meaning of the term: in Z[sqrt(-5)], 1+sqrt(-5) and 2 are coprime
> under your incorrect usage, but they are not coprime under correct usage.
They are coprime in the ring Z[sqrt(-5)].
It's strange to me that you'd miss something so obvious, as you could
also consider something like the ring of evens, where 2 is coprime to
6, in the ring of evens.
Apparently, you wish to use the knowledge that in a *higher* ring
1+sqrt(-5) and 2 share sqrt(2) as a factor, for the lesser ring.
But just as 2 is a factor of 6 in the ring of integers, but is NOT in
the ring of evens, it simply is the case that in the ring Z[sqrt(-5)]
1+sqrt(-5) and 2 are coprime, by the logical definition.
What appears to have happened is that mathematicians considered
various rings, and tried to find some general way to handle them,
rather than simply finding the object ring.
It'd be like someone fiddling around with the ring of evens, rather
than just integers. Sometimes they put in 3, other times they put in
7 with integers, when they really just need to put in 1, and be done.
> >However, by his definition of coprime, only one of the numbers is
> >coprime.
>
> "Coprime" to ->what<-?
>
> And prove it.
One of the a's is coprime to f, and in particular 5, with f=sqrt(5),
m=1.
The proof follows.
Consider, in the ring of algebraic integers,
P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f).
Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization
P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
where w_1 w_2 w_3 = f, and
b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),
and at m=0
P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
so two of the b's must equal 0, which means
P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)
which is
P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)
proving that w_1 w_2 must be coprime to 3, if f is coprime to 3, which
leaves
b_3 = 3, or a unit multiple of 3.
Essentially objections to how f^2 divides off now come down to
claiming that the w's are functions of m, but consider that w_1 w_2 =
1, when m=0, if f is coprime to 3.
But that was an arbitrary choice, so let f=3.
Now it can be shown that w_1 w_2 is coprime to 3 without regard to m.
That is, the w's can now be shown to all be constant with regard to m,
so they have the same value no matter what the value of m is, so they
are also constant
with f coprime to 3.
Introducing a_1, a_2, and a_3, that is seen by considering
P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f) =
(a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
with f=3, as then you have
P(m)/3^3 = (m^3 3^3 - 3m^2 (3) + m) x^3 -
(-1+mf^2 )x u^2 + u^3 =
(a_1/3 x + u)(a_2/3 x + uf)(a_3/3 x + u)
so I have a_1 = b_1/3, a_2 = b_2/3, a_3 = b_3/3 where the 3 divides
off as a constant so it is without a dependency on m.
Now looking at
P(m)/3^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
I can again check at m=0, to see
P(0)/3^2 = u^2 (b_3 w_1 w_2 x + 3u) = u^2(3x + 3u),
which forces w_3 = 3, or a unit multiple of 3.
Therefore, the factorization is
P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f =
(b_1 x + u)(b_2 x + u)(b_3 x + uf)
where you'll notice that the b's are algebraic integers with m=1,
f=sqrt(2), but that's a special case as generally they are not, which
shows a problem with the ring of algebraic integers.
> [.snip.]
>
> >> >>Lemma. Let R be the ring of all algebraic integers, and let a, b, c be
> >> >>any elements of R. If a and b are coprime to c, then a*b is coprime to
> >> >>c.
> >> >>
> >> >>Proof. We use the characterization of coprime valid for commutative
> >> >>rings with 1: a and b are coprime in R if and only if there exist x
> >> >>and y in R such that ax+by = 1.
> >> >
> >> >
> >> > By that definition only *one* of the a's is coprime to 5, but none of
> >> > them has a factor in common with 5 either.
> >>
> >> Do you even understand what he's saying? The application would be: If
> >> a_1 and a_2 are coprime to 5, then so is a_1*a_2. Extending it, if a_1,
> >> a_2, and a_3 are coprime to 5, then a_1*a_2 and a_3 are coprime to 5,
> >> which means a_1*a_2*a_3 is coprime to 5.
> >
> >By the definition he gave only ONE of the a's is coprime to 5.
>
> Prove it.
I've given the proof above. One of the a's is coprime by that
definition, while the others do not have non-unit factors in common
with 5 in the ring.
And in fact your example with Z[sqrt(-5)] outlines how, as similarly,
in that ring 1+sqrt(-5) is, by a *logical* definition*, coprime to 2.
However, by the definition you've been using, it would be that
1+sqrt(-5) and 2 do not share non-unit factors in the ring
Z[sqrt(-5)].
> >The others don't have non unit factors in common with 5, in the ring
> >of algebraic integers, but each have a factor that is sqrt(5) in a
> >higher ring.
>
> Which doesn't prove anything. If the others "don't have non-unit
> factors in common with 5" in the ring of all algebraic integers, then
> they would be coprime under the correct definition as well.
Actually, if the ring weren't flawed they *would* have non-unit
factors in common with 5.
> [.snip.]
>
> >> > For those who don't understand, consider that in the ring of evens,
> >> > which does not have 1, you can't use that definition of coprime that
> >> > Arturo Magidin gives, though it is, interestingly enough, true that in
> >> > fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is
> >> > not in the ring.
> >>
> >> Why do you make unnecessary side trips to bizarre rings?
> >
> >Arturo Magidin uses a special definition of coprime which depends on
> >certain assumptions which fail with the ring of algebraic integers,
>
> I use the STANDARD definition; it is you who uses a "SPECIAL"
> one.
>
> Please list, explicitly:
>
> 1. The "special definition";
I quote from above.
"It is a THEOREM that in a commutative ring with 1, two elements a and
b are coprime if and only if there exist x and y in R such that
ax+by=1."
> 2. The one you use;
I've given the following definition for coprime in the proper ring,
which is the ring of objects:
Two objects are coprime if they don't share a non-unit factor e.g. 2
is coprime to 3, while 2 and 4 share 2 as a non-unit factor.
> 3. The "certain assumptions" on which my supposedly special definition
> depends.
My guess is that you assume that all numbers that *should* be factors
in the ring of algebraic integers are units, when in fact the ring is
flawed in that certain numbers which *should* be units aren't in the
ring.
These numbers are multiplicative inverses of other numbers that *are*
in the ring of algebraic integers, but are not units, though they are
coprime to all non-unit algebraic integers.
The ring of algebraic integers is flawed in a fascinating way.
> 4. A proof that they fail in the ring of algebraic integers.
Well I just explained it, but I'll repeat that the problem is that
certain numbers which *should* be units in the ring, are not, and
their multiplicative inverses aren't even in the ring, though they
should be.
> It is not enough to say "the ring is really screwed up", because that
> is supposed to be your conclusion, not your hypothesis.
I've proven it repeatedly.
Rather than work to get to the truth, I've seen posters work to
convince, presumably to protect their *beliefs* about mathematics as a
discipline.
Their social preoccupation has been of interest to me.
My fear is that many "mathematicians" actually are participants in
what can be considered to be a dogmatic relgion, where rather than
challenging what they're taught, math students learn to accept from
authority.
However, science is about challenging what is known to make certain
that it fits with reality, not with giving your trust to particular
human beings.
"Trust, but verify", as Ronald Reagan said, in one of the best quotes
ever.
James Harris
Arturo Magidin
James Harris <jst...@msn.com> wrote:
>mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<bharf7$8p8$1...@agate.berkeley.edu>...
>> In article <3c65f87.03081...@posting.google.com>,
>> James Harris <jst...@msn.com> wrote:
>
><deleted>
>
>> >He uses a special definition of coprime, and assumes that at least two
>> >of the three numbers is coprime.
>>
>> No, I use the standard definition of coprime: two elements a and b are
>> coprime in R if and only if the ideals (a) and (b) are coprime, if and
>> only if there is no prime ideal of R which contains both (a) and (b).
>>
>> It is a THEOREM that in a commutative ring with 1, two elements a and
>> b are coprime if and only if there exist x and y in R such that
>> ax+by=1.
>
>That's fascinating. Why don't you give the theorem, as it must be in
>error for the ring of algebraic integers.
>
>And yes, I would like to see that theorem as I'd be fascinated to see
>some evidence of how deep the weird algebraic integer ring problem has
>gone in terms of what mathematicians believe to be true.
>
>Still my guess is that it depends on assertions about units, where you
>base your belief, once again, on assuming that a number that is not a
>unit in the ring of algebraic integers must be a non-unit factor in
>the ring of algebraic integers.
You are speaking nonsense. A "non unit factor" of ->what<-?
If you again mean, "a non-unit factor of itself", then there is no
belief, there is only mathematical truth: by definition, x is a factor
of y in the commutative ring R if and only there exists an element z
of R such that x*z=y. If a is an algebraic integer which is not a unit
in the ring of all algebraic integers, then since 1 is an algebraic
integer and a*1=a, it follows, necessarily, that a is a non-unit
factor of itself in the ring of all algebraic integers.
I'll include the definitions, since I doubt you know them.
DEF. Let R be a ring. A subset I of R is an "ideal of R" if and only
if:
(i) I is nonempty;
(ii) If x and y are in I, then so is x-y.
(iii) If x is in I and r is any element of R, then rx is in I and
xr is in I.
Note that if the ring is commutative, it is enough to check that rx is
in I.
Lemma 0. If I is an ideal of R, then 0 is in I; and if a,b are in I,
then a+b is in I.
Proof. I is nonempty, so there is an a in I. Since 0 is in R, 0*a=0
lies in I by (iii). So 0 is in I.
If a,b are in I, then -b=0-b lies in I by (ii). Since a,-b lie in I,
then a-(-b)=a+b lies in I by (ii). QED.
DEF. Let R be a commutative ring, and let a be an element of R. The "principal
ideal generated by a" is the collection { ra : r in R}. It is denoted
by (a).
Lemma 1. Let R be a commutative ring. Then (a) is an ideal of R.
Proof. Since R is a ring, it contains a 0. Therefore, 0*a=0 is in
(a), so (a) is nonempty. If ra and sa are in (a), then ra-sa=(r-s)a,
which by definition is in I. Finally, if ra is in (a), and s is any
element of R, then s*(ra) = (sr)a, which lies in (a). QED
DEF. Let R be a ring. An ideal I of R is said to be a "prime ideal" if
and only if it is not equal to R, and for any two elements x and y of
R, if x*y is in I, then either x is in I or y is in I.
DEF. Let R be a commutative ring. Two ideals I and J of R are said to
be "coprime" (in R) if and only if there is no prime ideal P of R which
contains both I and J.
DEF. Let R be a commutative ring. Two elements a and b of R are said
to be "coprime" (in R) if and only if the principal ideals (a) and (b)
are coprime in R.
DEF. Let R be a commutative ring. An ideal I of R is said to be a
"maximal ideal" if and only if I is not equal to R, and there is no
ideal J of R which properly contains I and is properly contained in R.
DEF. Let R be a commutative ring. Two ideals I and J of R are said to
be "comaximal" if and only if there does not exist a maximal ideal of
R which contains both I and J.
Lemma 2. Let I be an ideal of R, and let a be an element of R not in
I. Then the set
{ x + ra : x in I, r in R}
is an ideal of R, and moreover, it is the smallest ideal of R which
contains I and a.
Proof. Let S denote the set {x+ra : x in I, r in R}.
First, since I is an ideal, it is nonempty. Let x in I. Then x+0*a is
in S, so S is nonempty.
Assume that b,c are in S. Then b=x+ra for some x in I, r in R; and
c=y+sa for some y in I, s in R. Then
b-c = (x+ra) - (y+sa) = (x-y) + (r-s)a.
Since I is an ideal, x-y is in I; and r-s is in R. So this is an
element of S; thus, if b and c are in S, then so is b-c.
Finally, let y be any element of S and let r in R. Then we may write
y=x+sa with x in I and s in R. Then
r*y = r*(x+sa) = (r*x) + r*sa
= (rx) + (rs)*a.
Since I is an ideal, rx is in I; so this is an element of
S. Therefore, S is an ideal.
Now assume that J is any ideal which contains I and contains a. We
prove that it contains S. For let x in I be arbitrary and let r in
R. We want to show that x+ra is in J. Since J is an ideal, and a is in
J, it follows that ra is in J. Since J is an ideal, and ra is in J, it
follows that 0=0*ra is in J. Since 0 and ra are in J, it follows that
-ra = 0-ra is in J. Since x is in J and -ra is in J, it follows that
x-(-ra)=x+ra is in J. This proves that J contains S, and proves the
Lemma. QED.
THEOREM 3. Let R be a commutative ring with 1. If I is a maximal ideal
of R, and a is any element of R not in I, then there exists an element
x in I and an element r in R such that x+ra=1.
Proof. Let S={x+ra : x in I, r in R}. This set is an ideal, and
contains I. It also contains a, since a=0+1a. Therefore, the set
properly contains I. Since I is maximal, the only ideal which properly
contains I is R itself, so S=R. Since 1 is in R, we conclude that 1 is
in S, and therefore, 1 = x+ra for some x in I and r in R, as
claimed. QED.
THEOREM 4. Let R be a commutative ring with 1. If I is a maximal ideal
of R, then it is also a prime ideal of R.
Proof. Assume not. Then there exist elements a,b of R, none of them in
I, such that ab lies in I.
Since a is not in I, by Theorem 3 there exist x in I and r in R such
that 1 = x+ra. Multiplying by b we have
b = b*1 = b(x+ra) = bx + bra = (bx) + r(ba).
Now, x is in I, hence so is bx; likewise, ba is in I, hence so in
r*ba. Therefore, bx+r(ba) lies in I. But that means that b is in I,
contradicting the choice of a and b. Therefore, I must be a prime
ideal. QED
THEOREM 5. (Zorn's Lemma) If C is a nonempty partially ordered set with the
property that any chain is bounded above, then C has maximal elements.
Proof omitted. Known to be equivalent to the Axiom of Choice.
Lemma 6. Let R be a commutative ring with 1, and let I be an ideal. If
I contains 1, then I=R.
Proof. We know I is contained in R. To prove the reverse inclusion,
let x in R. We need to show it is in I. Since 1 in I and x in R, x*1=x
must lie in I, so x is in I. QED.
THEOREM 7. Let R be a ring with 1. If I is any ideal of R and is not
equal to R, then there exists a maximal ideal M of R which contains I.
Proof. Let S= { J ideal of R: J is not equal to R and J contains I}.
Order S by inclusion. Then I is in S, so S is nonempty. Let C be a
chain in S, that is, C is contained in S and C={J_k}, with the
property that for any indices m,n, either J_m is contained in J_n or
J_n is contained in J_m. If C is empty, I is an upper bound for
C. Assume C is nonempty. Let J be the set theoretic union of C. Then J
contains I, since each element of J contains I. So it is nonempty.
If a,b in J, then a in J_k and b in J_n for some k and n. Either J_k
is contained in J_n (and so J_n contains both a and b) or J_n
contained in J_k (so J_k contains both a and b). So assume without
loss of generality that J_n contains both a and b. Since J_n is an
element of S, it is an ideal, so a-b is in J_k, and so a-b is in J.
Finally, if a in J, and r in R, then a in J_k for some k, and since
this is an ideal, ra is in J_k, so ra is in J. Therefore, J is an
ideal of R which contains S.
J contains all elements of C, so it is a bound for C. We just need to
show that it lies in S; the only way in which it would not lie in S is
if it violated the condition "J is not equal to R". But that would
mean that J=R, so 1 is an element of J. Since J is the set-theoretic
union of the J_k, so 1 must be in some J_k. But that would imply, by
Lemma 6, that J_k=R; this contradicts the fact that J_k is an element
of S. The contradiction arose from assuming that R=J, so J is not
equal to R, so J lies in S.
In conclusion, S satisfies the hypothesis of Zorn's Lemma, and has
maximal elements. Let M be a maximal element of S.
That means that: (1) M is an ideal of R; (2) M is not equal to R; and
(3) M contains I [all this by virtue of being in S]. Finally, it also
means that there does not exist any ideal J of R, different from R,
which contains I, contains M, and is not equal to M [by virtue of
being maximal in S].
We need only show that M is maximal to prove the theorem. But if M is
not maximal, then there exists an ideal K of R, different from R,
which properly contains M. But then it would necessarily contain I,
and so be an element of S larger than M. This contradicts the
maximality of M in S, and so we conclude that M is a maximal ideal
that contains I. QED
THEOREM 8. Let R be a commutative ring with 1. Two ideals I and J are
coprime if and only if they are comaximal.
Proof. If I and J are coprime, and M is a maximal ideal, then M cannot
contain both I and J, for then M would also be a prime ideal
containing both I and J, which is impossible. So I and J are
comaximal.
Conversely, suppose that I and J are comaximal. If P is a prime ideal
containing both I and J, then by Theorem 7 there exists a maximal
ideal M containing P, and therefore containing I and J, contradicting
comaximality. So I and J are coprime. QED
THEOREM 9. Let R be a commutative ring with 1, and let a and b be two
elements of R. Then a and b are coprime in R if and only if there
exist x and y in R such that ax+by=1.
Proof. First, assume that there exists x and y in R such that
ax+by=1. Note that since R has a 1, a lies in (a) (being equal to
1*a), and b lies in (b). So if P is any ideal containing both (a)
and (b), then it contains both a and b as elements. Since a is in P,
then ax must be in P; since b is in P, then b*y is in P. Since ax and
by are both in P, we must have that ax+by is in P. But 1=ax+by, so 1
is in P. Therefore, P=R.
In particular, there is no ideal other than R which contains both (a)
and (b); therefore, there is no maximal ideal which contains both (a)
and (b); therefore, by definition, (a) and (b) are comaximal. By
Theorem 8, they are also coprime. Therefore, a and b are coprime in R.
Conversely, assume that a and b are coprime in R. That means that (a)
and (b) are coprime, hence comaximal by Theorem 8.
Let I = {ra + sb : r,s in R}. We claim that this is an ideal of R
which contains both (a) and (b). Indeed, it is an ideal, by the same
argument as Lemma 2: it is equal to the set of all x+sb with x in (a)
and s in R. It contains (a) because if ra is in (a), then
ra = ra+0*b lies in I; and it contains (b) because if sb is in (b),
then sb = 0*a+sb which is in I.
Therefore, I is an ideal which contains both (a) and (b). If I is not
equal to R, then by Theorem 7 there exists a maximal ideal M of R
which contains I; therefore, M contains (a) and (b). This contradicts
the comaximality of (a) and (b). The contradiction arises from
assuming that I is not equal to R, so we must conclude that
I=R. Therefore, 1 is an element of R, so 1 is in {ra+sb : r,s in
R}. Therefore, there exist x and y in R such that 1 = xa+yb, as
claimed. QED
Done.
>The problem is fascinating indeed, and easily leads to confusion, if
>you aren't careful to be *very* careful with your definitions and
>assumptions.
Which you are not.
>That is, you need to follow the math, not what you *think* is the
>truth.
>
>> What is the definition YOU are using?
>
>I've given the following definition for coprime in the proper ring,
>which is the ring of objects:
There is no "ring of objects". Your definition does not have a referent,
as I have pointed out many times.
This definition is in general weaker than the usual one, but is
equivalent to the usual one in the ring of all algebraic integers.
DEF. Let R be a commutative ring, a and b elements of R. We say that
"a is a factor of b (in R)" if and only if there exists c in R such that
ac=b.
DEF. Let R be a commutative ring with 1, and let u in R. Then u is a
unit in R if and only if there exists v in R such that u*v=1.
DEF. Let R be a commutative ring, a,b,c elements of R. We say that "c
is a common factor of a and b (in R)" if and only if c is a factor of
a in R and c is also a factor of b in R.
THEOREM. Let R be a ring commutative ring with 1, and let a and b in
R. If a and b are coprime in R, then any common factor of a and b in R
is a unit.
Proof. Since a and b are coprime in R, there exist x,y in R such that
ax+by=1. Let c in R be a common factor of a and b.
Then there exists d in R such that cd=a; and there exists e in R such
that ce=b. Therefore, cdx=ax, and cey=by. Therefore,
1 = ax+by = cdx + cey = c(dx+ey).
Since R is a ring, dx and ey are in R, and so is dx+ey. Therefore,
dx+ey is an element of R which multiplied by c yields 1. Therefore, c
is a unit in R. QED
EXAMPLE. A commutative ring with 1 R in which two elements have no
common factor other than units, but are not coprime.
Proof. Let R = Z[sqrt(-5)], a = 2, b= 1+sqrt(-5).
First, we show that a and b are not coprime. For assume that we had
x,y in R such that ax+by = 1. Write x = r+s*sqrt(-5), y =
t+u*sqrt(-5) with r,s,t,u integers. Then
1 = 2x+(1+sqrt(-5))y
= 2r + 2s*sqrt(-5) + t + u*sqrt(-5) -5u +t*sqrt(-5)
= (2r + t - 5u) + (2s+u+t)*sqrt(-5).
Therefore, 2r+t-5u = 1
2s+u+t = 0
Substituting t=-2s-u, we have
1 = 2r -2s -u -5u
= 2r - 2s - 6u.
But since r,s,u are integers, that means that 2r-2s-6u is an even
integer, and therefore cannot equal 1. Therefore, a and b are not
coprime in R.
To prove that they have no common factors, consider the function
N:Z[sqrt(-5)]->Z given by
N(a+b*sqrt(-5)) = a^2 + 5b^2, where a and b are integers.
Then it is easy to verify by direct calculation that for all x,y in Z[sqrt(-5)]
N(xy) = N(x)N(y).
Assume that x is a factor of y in Z[sqrt(-5)]. Then there exists z in
Z[sqrt(-5)] such that xz=y. Therefore, N(y) = N(xz) =
N(x)N(z). Therefore, the integer N(y) is a product of the integers
N(x) and N(z), so N(x) is a factor of N(y) IN THE INTEGERS.
That is: if x is a factor of y in Z[sqrt(-5)], then N(x) is a factor
of N(y) in Z.
So, assume that a+b*sqrt(-5) is a common factor of 2 and
1+sqrt(-5). Then N(a+bsqrt(-5)) = a^2 + 5b^2 must be a factor, IN THE
INTEGERS, of N(2) = 4 and of N(1+sqrt(-5)) = 1+1*5 = 6.
The only possibilities are that N(a+b*sqrt(-5)) = 2 or 1 (note N(x) is
always nonzero).
But 2 = a^2+5b^2 has no solution with a and b integers, so we must
have N(a+b*sqrt(-5)) = 1. The only solutions to
1 = N(a+b*sqrt(-5)) = a^2 + 5b^2 are b=0, a=1; or b=0, a=-1. So either
a+b*sqrt(-5)=1, hence is a unit; or a+b*sqrt(-5)=-1, hence is a
unit. This proves that
(1) 2 and 1+sqrt(-5) are not coprime in Z[sqrt(-5)]; but
(2) 2 and 1+sqrt(-5) have no non unit common factors in Z[sqrt(-5)].
This proves that in general the two notions are not equivalent; since
coprime implies no common factors other than units, but the converse
does not hold, "no common factors other than units" is a weaker
proposition than coprime, in general.
The only thing that would remain is to prove that the two notions are
equivalent in the ring of all algebraic integers. This is a deep
theorem due to Dedekind and Kronecker, and uses the finiteness of the
class number. It has been sketched before, but here are the broad
strokes:
First, let R be the ring of all algebraic integers. For any a and b in
R, there exists c in R such that
(a,b) = {xa + yb : x, y in R} = (c)
To prove this, let A be the ring of integers of Q(a,b). Then there
exists a positive integer k such that (a,b)^k = (d) for some d in A
(finiteness of the class number). Let A' be the ring of integers of
Q(a,b,d^{1/k}), where d^{1/k} is any of the complex k-th roots of
d. In A', we have (a,b)^k = (d) = (d^{1/k})^k, so by unique
factorization of ideals in rings of integers, we have
(a,b)=(d^{1/k}). From here, it follows that the two sets above are
equal, so letting c=d^{1/k} gives c.
Let a and b be any two algebraic integers, and assume that a and b
have no common non-unit factors in the ring of all algebraic
integers. Let c be an algebraic integer such that (a,b)=(c). Then c
is a common factor of a and b: for a is in (a,b), and so a is in
(c)={rc: r in R}; that means, there exists r in R such that
rc=a. Likewise, b is in (a,b), hence in (c), and so there exists s in
R such that cs=b. So c is a common factor of both a and b. Since a and
b have no common factors other than units, that means that c is a unit
in the ring of algebraic integers. That means that there exists v in R
such that vc=1. But c is in the ideal (c), and v is in R, so vc=1 is
in (c). Since (c) contains 1, it follows that (c)=R.
Therefore, (a,b) = {xa+yb : x,y in R} = (c) = R, so 1 is in
(a,b). Therefore, there exist x and y in R such that ax+by
=1. Therefore, a and b are coprime in R, as claimed.
[.rest deleted.]
>> In the ring of all algebraic integers, the two are
>> EQUIVALENT. However, your definition is WEAKER than the correct
>> meaning of the term: in Z[sqrt(-5)], 1+sqrt(-5) and 2 are coprime
>> under your incorrect usage, but they are not coprime under correct usage.
>
>They are coprime in the ring Z[sqrt(-5)].
Not under the correct usage.
>It's strange to me that you'd miss something so obvious, as you could
>also consider something like the ring of evens, where 2 is coprime to
>6, in the ring of evens.
They are coprime under the usual definition, and under your
bastardization thereof. However, in the ring Z[sqrt(-5)], they are
"coprime" under your bastardization, but not under the correct usage.
[.snip.]
>> >However, by his definition of coprime, only one of the numbers is
>> >coprime.
>>
>> "Coprime" to ->what<-?
>>
>> And prove it.
>
>One of the a's is coprime to f, and in particular 5, with f=sqrt(5),
>m=1.
>
>The proof follows.
>
>Consider, in the ring of algebraic integers,
>
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>
> 3(-1+mf^2 )x u^2 + u^3 f).
>
You have no u in your assertions before; what is u?
>Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization
>
> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
You have not proven that b1, b2, b3, u, w1, w2, w3 are algebraic
integers.
>where w_1 w_2 w_3 = f, and
>
> b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),
>
>and at m=0
>
> P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
>
>so two of the b's must equal 0, which means
WHEN m = 0.
> P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)
>
>which is
>
> P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)
WHEN m = 0.
>proving that w_1 w_2 must be coprime to 3
YOu have not proven that w1*w2 is an algebraic integer, so this
statement is unsupported right now.
>if f is coprime to 3, which
>leaves
>b_3 = 3, or a unit multiple of 3.
WHEN m = 0.
You have, however, not proven that, under "my definition", they are
coprime.
>Essentially objections to how f^2 divides off now come down to
>claiming that the w's are functions of m, but consider that w_1 w_2 =
>1, when m=0, if f is coprime to 3.
This is nonsense and is based on your misunderstanding of obejctions.
>But that was an arbitrary choice, so let f=3.
WHEN m = 0.
>Now it can be shown that w_1 w_2 is coprime to 3 without regard to m.
Please do so, using the correct definition of "coprime", since you
claim that "under my definition" you will do so. Be sure to prove that
w_1*w_2 is an algebraic integer while you do so.
>That is, the w's can now be shown to all be constant with regard to m,
Please do so.
>so they have the same value no matter what the value of m is, so they
>are also constant
>with f coprime to 3.
Please provide a complete proof. See above for what a real proof looks like.
>Introducing a_1, a_2, and a_3, that is seen by considering
>
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>
> 3(-1+mf^2 )x u^2 + u^3 f) =
>
> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
You have not proven that a1, a2, a3 are algebraic integers.
>with f=3, as then you have
>
> P(m)/3^3 = (m^3 3^3 - 3m^2 (3) + m) x^3 -
>
> (-1+mf^2 )x u^2 + u^3 =
>
> (a_1/3 x + u)(a_2/3 x + uf)(a_3/3 x + u)
>
>so I have a_1 = b_1/3, a_2 = b_2/3, a_3 = b_3/3 where the 3 divides
>off as a constant so it is without a dependency on m.
when f=3. You have not shown that a_1/3, a_2/3, a_3/3 are algebraic integers.
>Now looking at
>
> P(m)/3^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
>
>I can again check at m=0, to see
>
> P(0)/3^2 = u^2 (b_3 w_1 w_2 x + 3u) = u^2(3x + 3u),
>
>which forces w_3 = 3, or a unit multiple of 3.
When m=0.
>Therefore, the factorization is
>
> P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>
> 3(-1+mf^2 )x u^2 + u^3 f =
>
> (b_1 x + u)(b_2 x + u)(b_3 x + uf)
>
>where you'll notice that the b's are algebraic integers with m=1,
I don't notice that; I assume, by the way, that you screwed up above
and when you wrote
> a_1 = b_1/3, a_2 = b_2/3, a_3 = b_3/3
you meant
> b_1 = a_1/3, b_2 = a_2/3, b_3 = a_3/3.
>f=sqrt(2), but that's a special case as generally they are not, which
>shows a problem with the ring of algebraic integers.
YOu have analized a number of special cases, and said a lot of things,
most of which are not supported. In addition, you CLAIMED that you
would prove that, under "my" definition (i.e., the standard
definition), the ai would be shown to be coprime to 5 for the
polynomial in question, 65x^3 - 12x + 1.
I do not see the values of r_i and s_i such that r_i*a_i + s_i*5 = 1,
which is the definition you claimed you were going to use.
So all you did was waste space and my time.
[.snip.]
>> >By the definition he gave only ONE of the a's is coprime to 5.
>>
>> Prove it.
>
>I've given the proof above. One of the a's is coprime by that
>definition,
I do not see the elements r and s that make r*a+s5 = 1. That's what
the definition says, after all. PROVE IT.
>while the others do not have non-unit factors in common
>with 5 in the ring.
You have not proven anything.
>And in fact your example with Z[sqrt(-5)] outlines how, as similarly,
>in that ring 1+sqrt(-5) is, by a *logical* definition*, coprime to 2.
The STANDARD DEFINITION gives that 1+sqrt(-5) and 2 are NOT coprime in
Z[sqrt(-5)]. I've proven it.
>However, by the definition you've been using, it would be that
>1+sqrt(-5) and 2 do not share non-unit factors in the ring
>Z[sqrt(-5)].
That is true regardless of whether you use "my" definition of
"coprime" (i.e., the correct one), or "your" definition of coprime
(i.e., the bastardization).
The point is that this property does not imply coprimality, using the
correct definition.
>> >The others don't have non unit factors in common with 5, in the ring
>> >of algebraic integers, but each have a factor that is sqrt(5) in a
>> >higher ring.
>>
>> Which doesn't prove anything. If the others "don't have non-unit
>> factors in common with 5" in the ring of all algebraic integers, then
>> they would be coprime under the correct definition as well.
>
>Actually, if the ring weren't flawed they *would* have non-unit
>factors in common with 5.
You are assuming your conclusion. That's called a ciruclar argument.
>> >Arturo Magidin uses a special definition of coprime which depends on
>> >certain assumptions which fail with the ring of algebraic integers,
>>
>> I use the STANDARD definition; it is you who uses a "SPECIAL"
>> one.
>>
>> Please list, explicitly:
>>
>> 1. The "special definition";
>
>I quote from above.
>
>"It is a THEOREM that in a commutative ring with 1, two elements a and
> b are coprime if and only if there exist x and y in R such that
> ax+by=1."
You see the big capitalized word "THEOREM"?
You did not give the "special definition" I supposedly use.
>> 2. The one you use;
>
>I've given the following definition for coprime in the proper ring,
>which is the ring of objects:
There is no ring of objects. Your definition is incoherent as written,
and has no referent when fixed.
>Two objects are coprime if they don't share a non-unit factor e.g. 2
>is coprime to 3, while 2 and 4 share 2 as a non-unit factor.
This is not the correct definition of "coprime".
>
>> 3. The "certain assumptions" on which my supposedly special definition
>> depends.
>
>My guess
You ->guess<-?
But you made a very definitive ASSERTION above. Are you trying to say
that when you asserted that I was depending on some assertions, you
did not KNOW that to be the case, you were merely ->guessing<-?
Is that how you work? [rhetorical question: yes, it is how you work]
>is that you assume that all numbers that *should* be factors
>in the ring of algebraic integers are units, when in fact the ring is
>flawed in that certain numbers which *should* be units aren't in the
>ring.
All this "should" really means "numbers which I WOULD REALLY, REALLY,
REALLY like to have the properties I want so my proof wouldn't be
nonsense."
[.snip.]
>These numbers are multiplicative inverses of other numbers that *are*
>in the ring of algebraic integers, but are not units, though they are
>coprime to all non-unit algebraic integers.
If the multiplicative inverses are NOT algebraic integers, then you
cannot talk about them being coprime to algebraic integers in the ring
of algebraic integers. If they are multiplicative inverses of
something in your ring, then they are units, and therefore coprime to
EVERYTHING in the ring. In short, what you wrote is a lot of empty
verbiage.
>The ring of algebraic integers is flawed in a fascinating way.
So you claim.
>> 4. A proof that they fail in the ring of algebraic integers.
>
>Well I just explained it, but I'll repeat that the problem is that
>certain numbers which *should* be units in the ring, are not, and
>their multiplicative inverses aren't even in the ring, though they
>should be.
Translation: numbers that you ->fervently wish<- were there are not,
and that's a problem. For you.
>> It is not enough to say "the ring is really screwed up", because that
>> is supposed to be your conclusion, not your hypothesis.
>
>I've proven it repeatedly.
You've given a lot of verbiage, but you have proven nothing.
C. Bond
> ...I'd be fascinated to see...
and
> ...The problem is fascinating indeed, ...
and
> ...is flawed in a fascinating way...
James. Enough with the "fascinating". Please drop the "Yup.", "Um.", "odd", "amazing", etc. as well. If
your vocabulary needs work, try remedial English.
[snip]
> Well I just explained it, but I'll repeat that the problem is that
> certain numbers which *should* be units in the ring, are not, and
> their multiplicative inverses aren't even in the ring, though they
> should be.
Fine. Name one. Give us a number which *should* be a unit in the ring of algebraic integers, but which is
not. Just one. Pretty please! For God's sake man, name just ONE numeric quantity which supports your
claim!
William Hale
> In article <3c65f87.03081...@posting.google.com>,
> James Harris <jst...@msn.com> wrote:
> >mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<bharf7$8p8$1...@agate.berkeley.edu>...
> >> In article <3c65f87.03081...@posting.google.com>,
> >> James Harris <jst...@msn.com> wrote:
> >It's strange to me that you'd miss something so obvious, as you could
> >also consider something like the ring of evens, where 2 is coprime to
> >6, in the ring of evens.
>
> They are coprime under the usual definition, and under your
> bastardization thereof. However, in the ring Z[sqrt(-5)], they are
> "coprime" under your bastardization, but not under the correct usage.
This seems to be a harsh characterization of Harris's definition
of "coprime".
Herstein defines two numbers to be "relatively prime" (ie, coprime)
if their greatest common divisor is a unit in the given ring. But,
this definition is applied only to Euclidean rings.
This is the definition I am used to, so I don't object that much
to Harris's definition.
Furthermore, this definition shows where the terminology "coprime"
originated, whereas your definition is more removed from "prime".
As you point out, Harris's definition in more general rings may
not provide the "usual" theorems one wants, so the definition
of "coprime" needs to be modified or generalized (which was
done as you stated).
In summary, I don't think that Harris's definition of "coprime"
is at all non-standard, and if he wants to use it, I see no
problem with it.
-- Bill Hale
Arturo Magidin
William Hale <ha...@tulane.edu> wrote:
>mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<bhbev8$m80$1...@agate.berkeley.edu>...
>> In article <3c65f87.03081...@posting.google.com>,
>> James Harris <jst...@msn.com> wrote:
>> >mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<bharf7$8p8$1...@agate.berkeley.edu>...
>> >> In article <3c65f87.03081...@posting.google.com>,
>> >> James Harris <jst...@msn.com> wrote:
>[cut]
>> >It's strange to me that you'd miss something so obvious, as you could
>> >also consider something like the ring of evens, where 2 is coprime to
>> >6, in the ring of evens.
>>
>> They are coprime under the usual definition, and under your
>> bastardization thereof. However, in the ring Z[sqrt(-5)], they are
>> "coprime" under your bastardization, but not under the correct usage.
>
>This seems to be a harsh characterization of Harris's definition
>of "coprime".
>
>Herstein defines two numbers to be "relatively prime" (ie, coprime)
>if their greatest common divisor is a unit in the given ring. But,
>this definition is applied only to Euclidean rings.
In Euclidean rings (in fact, in any PID or Bezout domain), the two
notions are equivalent.
>This is the definition I am used to, so I don't object that much
>to Harris's definition.
>
>Furthermore, this definition shows where the terminology "coprime"
>originated, whereas your definition is more removed from "prime".
I disagree; the definition "coprime" as "have no common divisors other
than units" has ->no<- reference whatsoever to primes. Whereas the
definition "coprime" as "there is no prime ideal which contains both"
is directly related to primes.
But I guess that's neither here nor there.
James Harris
> In article <bda6703.03081...@posting.google.com>,
> William Hale <ha...@tulane.edu> wrote:
> >mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<bhbev8$m80$1...@agate.berkeley.edu>...
> >> In article <3c65f87.03081...@posting.google.com>,
> >> James Harris <jst...@msn.com> wrote:
> >> >mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<bharf7$8p8$1...@agate.berkeley.edu>...
> >> >> In article <3c65f87.03081...@posting.google.com>,
> >> >> James Harris <jst...@msn.com> wrote:
> [cut]
> >> >It's strange to me that you'd miss something so obvious, as you could
> >> >also consider something like the ring of evens, where 2 is coprime to
> >> >6, in the ring of evens.
> >>
> >> They are coprime under the usual definition, and under your
> >> bastardization thereof. However, in the ring Z[sqrt(-5)], they are
> >> "coprime" under your bastardization, but not under the correct usage.
> >
> >This seems to be a harsh characterization of Harris's definition
> >of "coprime".
I agree with William Hale.
> >Herstein defines two numbers to be "relatively prime" (ie, coprime)
> >if their greatest common divisor is a unit in the given ring. But,
> >this definition is applied only to Euclidean rings.
That's interesting, and it makes logical sense.
> In Euclidean rings (in fact, in any PID or Bezout domain), the two
> notions are equivalent.
Well at issue is the question of whether or not the ring of algebraic
integers is flawed in a fascinating way. Your defense of the ring
involves the supposition that you can define "coprime" in a way that
I've shown presupposes that the ring does not have a problem.
William Hale has taken a more defensible position based on his
experience with the term "relatively prime" and you are apparently
trying to attack his view.
> >This is the definition I am used to, so I don't object that much
> >to Harris's definition.
> >
> >Furthermore, this definition shows where the terminology "coprime"
> >originated, whereas your definition is more removed from "prime".
>
> I disagree; the definition "coprime" as "have no common divisors other
> than units" has ->no<- reference whatsoever to primes. Whereas the
> definition "coprime" as "there is no prime ideal which contains both"
> is directly related to primes.
However, you presuppose that certain numbers *are* prime ideals,
trying to use your terminology there, when in fact, I say *provably*
they are not, but in fact there is a HIGHER RING where they share
non-unit factors.
(Maybe I should look up "prime ideals" but the effort doesn't seem to
be worth it. I assume some poster will get excited and post something
if I messed up.)
The example that you yourself brought up which I'll use as an analogy
is 1+sqrt(-5) in the ring Z[sqrt(-5)], which is coprime ***in the
ring*** to 2, though it shares a factor of sqrt(2) in the ring of
algebraic integers.
Now I'm saying that a similar situation can be found with the ring of
algebraic integers, which you are trying to dispute, but I've found
that you've used ***definitions*** rather than mathematical logic,
often in your posts.
> But I guess that's neither here nor there.
Only if you wish to convince rather than get to the truth.
Mathematical research is not about how many people agree with you,
based on your ability to convince, but in getting to the truth.
In my view, discussions are not head-to-head debates where one
position is to be defended at all costs, but ways to work your way to
the actual mathematical truth.
Using definitions to convince is a bad idea, as the proof is out
there.
James Harris
David C. Ullrich
No, that's not at issue. (Hint: If I started posting to sci.astronomy
insisting that the moon was made of green cheese it would not
follow that the question of whether the moon was made of
green cheese was "at issue"...)
>Your defense of the ring
>involves the supposition that you can define "coprime" in a way that
>I've shown presupposes that the ring does not have a problem.
>
>William Hale has taken a more defensible position based on his
>experience with the term "relatively prime" and you are apparently
>trying to attack his view.
>
>> >This is the definition I am used to, so I don't object that much
>> >to Harris's definition.
>> >
>> >Furthermore, this definition shows where the terminology "coprime"
>> >originated, whereas your definition is more removed from "prime".
>>
>> I disagree; the definition "coprime" as "have no common divisors other
>> than units" has ->no<- reference whatsoever to primes. Whereas the
>> definition "coprime" as "there is no prime ideal which contains both"
>> is directly related to primes.
>
>However, you presuppose that certain numbers *are* prime ideals,
<giggle> no, he's not saying that certain _numbers_ are prime
ideals. An ideal in a ring is not an element of the ring, it is a
subset of the ring satisfying certain conditions.
>trying to use your terminology there, when in fact, I say *provably*
>they are not, but in fact there is a HIGHER RING where they share
>non-unit factors.
>
>(Maybe I should look up "prime ideals" but the effort doesn't seem to
>be worth it.
<giggle>
> I assume some poster will get excited and post something
>if I messed up.)
>
>The example that you yourself brought up which I'll use as an analogy
>is 1+sqrt(-5) in the ring Z[sqrt(-5)], which is coprime ***in the
>ring*** to 2, though it shares a factor of sqrt(2) in the ring of
>algebraic integers.
>
>Now I'm saying that a similar situation can be found with the ring of
>algebraic integers, which you are trying to dispute, but I've found
>that you've used ***definitions*** rather than mathematical logic,
>often in your posts.
>
>> But I guess that's neither here nor there.
>
>Only if you wish to convince rather than get to the truth.
>
>Mathematical research is not about how many people agree with you,
>based on your ability to convince, but in getting to the truth.
>
>In my view, discussions are not head-to-head debates where one
>position is to be defended at all costs, but ways to work your way to
>the actual mathematical truth.
>
>Using definitions to convince is a bad idea, as the proof is out
>there.
It's astonishing that you don't realize how ridiculous that last
statement sounds.
>James Harris
************************
David C. Ullrich
C. Bond
[snip]
> Using definitions to convince is a bad idea,
Definitions are not used to convince, but to identify what you are talking about.
> as the proof is out
> there.
Just not in your website.
Arturo Magidin
[.snip.]
>Well at issue is the question of whether or not the ring of algebraic
>integers is flawed in a fascinating way. Your defense of the ring
>involves the supposition that you can define "coprime" in a way that
>I've shown presupposes that the ring does not have a problem.
There is no supposition. There are no problems. The definition I gave
is prefectly valid for any commutative ring. I posted the definition
when you requested a proof of the equivalence, but apparently you did
not bother to look for it.
>William Hale has taken a more defensible position based on his
>experience with the term "relatively prime" and you are apparently
>trying to attack his view.
Did you notice when he said that the definition he was familiar with
was only given for "Euclidean rings"?
Do you know what an Euclidean ring is?
Did you know that the algebraic integers are ->not<- a Euclidean ring?
And neither is 2Z?
>> >Furthermore, this definition shows where the terminology "coprime"
>> >originated, whereas your definition is more removed from "prime".
>>
>> I disagree; the definition "coprime" as "have no common divisors other
>> than units" has ->no<- reference whatsoever to primes. Whereas the
>> definition "coprime" as "there is no prime ideal which contains both"
>> is directly related to primes.
>
>However, you presuppose that certain numbers *are* prime ideals,
James, you admit below that you do not know what "prime ideals"
are. How could you possibly make this statement with a straight face
if you do not know what "prime ideals" are?
No, I do not "presuppose that certain numbers *are* prime ideals."
Ideals, and in particular prime ideals, are not ->numbers<-, they are
subsets of the ring with certain properties.
>trying to use your terminology there, when in fact, I say *provably*
>they are not, but in fact there is a HIGHER RING where they share
>non-unit factors.
How can you say that "numbers are not prime ideals" is something you
can PROVE, when you admit below that you do not know what "prime
ideal" means?
What does that do to your credibility, and the image of your grasp of
what you are talking about?
>(Maybe I should look up "prime ideals" but the effort doesn't seem to
>be worth it. I assume some poster will get excited and post something
>if I messed up.)
>
>The example that you yourself brought up which I'll use as an analogy
>is 1+sqrt(-5) in the ring Z[sqrt(-5)], which is coprime ***in the
>ring*** to 2, though it shares a factor of sqrt(2) in the ring of
>algebraic integers.
No, they are ->not<- coprime in the ring, they simply have no common
factors other than units. Z[sqrt(-5)] is also not a Euclidean domain,
so the definition given by William Hale does not apply (and in a
Euclidean domain, the two definitions are equivalent).
Note, as well, as I pointed out to you already two times, that this
definition of "coprime" as "have no common factors other than units"
is USELESS for your purposes, because what you want to use your
conclusion for is:
"If a*b=0 (mod c) and a and c are coprime, then b=0 (mod c)."
This statement is always true if you use the standard definition of
"coprime" for a commutative ring with 1 (no prime ideal contains both
(a) and (c) ), but it is FALSE in many rings if you use the definition
"have no common factors other than units".
In the example you have now parrotted, we have
(1+sqrt(-5))*(1-sqrt(-5)) = 0 (mod 2),
but you do NOT have (1-sqrt(-5))=0 (mod 2) even though 2 and
1+sqrt(-5) have no common factors other than units.
>Now I'm saying that a similar situation can be found with the ring of
>algebraic integers,
No, you cannot. You are in error.
>which you are trying to dispute, but I've found
>that you've used ***definitions*** rather than mathematical logic,
>often in your posts.
Definitions are ->part<- of mathematical logic. Your insistence that
you can do mathematics without definitions is only evidence that you
have no idea what mathematics is, and cannot recognize logic.
Will Twentyman
Based on his comments and arguments, this sounds about right. It's too
bad that *he* hasn't responded to confirm/deny this.
>
> The interesting point, of course, is that the definition James is
> using, if it were not equivalent to the usual one, would be useless
> for his purposes. James needs the coprime property in order to remove
> factors in a congruence, but that step is not valid if you only have
> "have no nonunit common factors". You need coprime in the usual sense,
> with its equivalent statement that there are x and y such that
> ax+by=1. (If the two are equivalent in the ring in question, of
> course, there is no problem; but they are not always equivalent).
>
> So, in the end, the definition which James is using is not the one he
> should be using for his purposes, unless he can prove that in the ring
> he is working on it is equivalent to the usual one. Never mind that
> there is no "ring of objects" as he has defined it.
>
> [.snip.]
He has yet to indicate something I can interpret as a clear way to
determine whether any particular object is in his "ring". He has (sort
of) defined the properties he wants the things in it to have, but not
how to identify them.
>>Which ring is that? If you mean your "Ring of Objects", please explain
>>why sqrt(5) is a factor there and not in the algebraic integers.
>
>
> There is no "Ring of Objects" according to his definition. Even
> allowing for the possibility that "number" means "complex number" or
> "algebraic number", there is no unique ring determined by the
> conditions he gives, yet he assumes there is one and only one such ring.
That's what I thought. Since it includes the algebraic integers the
only thing I would claim with any kind of confidence is that it is a
subring of the complex numbers, but for all I can tell it's the
algebraic numbers.
> Arturo Magidin
> mag...@math.berkeley.edu
Arturo Magidin
Will Twentyman <wtwen...@read.my.sig> wrote:
>Arturo Magidin wrote:
[.snip.]
>> There is no "Ring of Objects" according to his definition. Even
>> allowing for the possibility that "number" means "complex number" or
>> "algebraic number", there is no unique ring determined by the
>> conditions he gives, yet he assumes there is one and only one such ring.
>
>That's what I thought. Since it includes the algebraic integers the
>only thing I would claim with any kind of confidence is that it is a
>subring of the complex numbers, but for all I can tell it's the
>algebraic numbers.
Well, his definition does not necessarily include all the algebraic
integers; but it certainly excludes the possibility that the ring in
question is the entire algebraic numbers. Right now, the definition on
his webpage reads:
"The Object Ring is a commutative ring that includes all numbers such
that -1 and 1 are the only members that are both a unit and an
integer, where no non-unit member is a factor of any two integers that
are coprime."
Correcting "numbers" to "algebraic integers", and dropping the final
clause (it is empty and true in any ring, if we assume that "coprime"
means "coprime in the ring of integers"), then the first problem is
that he is referring to it as "THE" ring of Objects, but says it is
"a" commutative ring (I note he corrected 'set' to 'commutative ring',
finally).
So we would have:
"The Object Ring is a subring of the algebraic numbers such that 1
and -1 are the only rational integers which are units."
Now, there are many such rings. Any subring of the algebraic integers
will do, but there are also many others that will work, not all of
them in the algebraic integers. (The following was suggested by Bill
Dubuque). If you take any number field K, and a rational prime (p)
which factors as a product of distinct primes, at least one of which
is principal, you can invert the generator of that one by itself and
get a ring that satisfies the condition. So, for example, if you take
the ideal (7) in Z[sqrt(2)], which splits as a product of two distinct
prime ideals, (7) = (3+sqrt(2))(3-sqrt(2)), then you can consider the
ring A[1/(3+sqrt(2))], where A is the ring of all algebraic integers;
and you can also consider the ring A[1/(3-sqrt(2))], so presumably
both 1/(3+sqrt(2)) and 1/(3-sqrt(2)) are "objects", but any ring
containing both will contain 1/7, which is disallowed. So there is no
referent.
In addition, Bill Dubuque also pointed out that some arguments James
has made indicate that he wants the "ring of objects" (whatever it may
turn out to be) to be closed under conjugates. That is, if r is an
"object, and f(x) is the monic irreducible polynomial in Q[x] with r
as a root, then every complex root of f(x) is also an object. But in
that case, the only rings that will satisfy his condition on units are
the subrings of the algebraic integers, and he has therefore not
defined anything new.
Will Twentyman
Arturo already posted it. Look up from this post 4 levels. It was in
all that stuff that you probably skimmed over. Considering the large
amount of material he posted, you would probably be better off reading
and understanding it than ignoring it.
> Using definitions to convince is a bad idea, as the proof is out
> there.
How are we supposed to talk meaninfully if we aren't allowed to define
things? Would you be happy if I argued with you while using a different
definition of the term "algebraic integer" from yours?
Here's the problem. If you want to use different definitions, fine.
But the results you generate (if valid) will NOT apply to the
conventional definitions. Using your personal definitions that do not
correspond to the rest of the mathematical community's means your
results will need to be translated before it can be determined whether
your results have the same meaning that they sound like they do. Along
the way, defining all your terms would help.
You have mentioned that 2 and 3 are coprime in your "Ring of Objects".
Is 1/2 in it? If so, they are coprime because 2 is a unit. If not,
perhaps because there is a lack of factors. *Why* are they coprime? I
can't tell from your definition of coprime because I'm not sure what's
*in* the Ring of Objects.
Will Twentyman
> In article <3f3a8c08$1_4@newsfeed>,
> Will Twentyman <wtwen...@read.my.sig> wrote:
>
>>Arturo Magidin wrote:
>
>
> [.snip.]
>
>
>>>There is no "Ring of Objects" according to his definition. Even
>>>allowing for the possibility that "number" means "complex number" or
>>>"algebraic number", there is no unique ring determined by the
>>>conditions he gives, yet he assumes there is one and only one such ring.
>>
>>That's what I thought. Since it includes the algebraic integers the
>>only thing I would claim with any kind of confidence is that it is a
>>subring of the complex numbers, but for all I can tell it's the
>>algebraic numbers.
>
>
> Well, his definition does not necessarily include all the algebraic
> integers;
At one point I thought he had specified that in did. Interesting. I
know he keeps claiming that it is a "higher ring" so it seems that he
intends for it to include all the algebraic integers.
> but it certainly excludes the possibility that the ring in
> question is the entire algebraic numbers. Right now, the definition on
> his webpage reads:
>
> "The Object Ring is a commutative ring that includes all numbers such
> that -1 and 1 are the only members that are both a unit and an
> integer, where no non-unit member is a factor of any two integers that
> are coprime."
>
> Correcting "numbers" to "algebraic integers", and dropping the final
> clause (it is empty and true in any ring, if we assume that "coprime"
> means "coprime in the ring of integers"), then the first problem is
> that he is referring to it as "THE" ring of Objects, but says it is
> "a" commutative ring (I note he corrected 'set' to 'commutative ring',
> finally).
I suspect (see above) that he meant "numbers". I suspect he would
prefer to keep it a subring of the algebraic numbers with the algebraic
integers as a subring of it.
So if he insists on it having the algebraic integers, then the "Ring of
Objects" *is* the algebraic integers, and it still doesn't have the
properties he wants it to have.
Arturo Magidin
Will Twentyman <wtwen...@read.my.sig> wrote:
>Arturo Magidin wrote:
[.snip.]
>So if he insists on it having the algebraic integers, then the "Ring of
>Objects" *is* the algebraic integers, and it still doesn't have the
>properties he wants it to have.
Well, if he insists on having all the algebraic integers, and the ring
having the property he listed (no rational integer other than 1 and -1
is invertible), AND be closed under conjugates, then it has to be the
algebraic integers. If all he does is drop the "closed under
conjugates", then there is no "largest" ring which satisfies his
definition, and there are many such rings, and he needs to specify
which it is he wants before anything can even begin to make sense.
Of course, he can always just change the definition again next time he
has a revelation and "discovers" there's a mistake. I'm still
wondering just what exactly was the problem he "found" with the previous
definition, and whether it had anything to do with the multiple
problems that many people pointed out.
And, since it is unclear just what properties he wants, well, your
guess is as good as mine. For instance, one of his claims for needing
objects is that the complete factorization of polynomials into linear
terms is not necessarily possible over the algebraic integers, though
he has not provided a counterexample nor found any errors in the
proofs or references provided. He may get a ring where his claims
about common factors hold, but then he would need to prove that having
no common factor implies the cancellation result he wants.
And, in any case, one still has to wonder at the amazing resilience of
his "proof", when the fundamental concept can change and yet no word
needs to be changed in the proof to accommodate this change. I guess
all the changes go into that big gap where he asserts a factorization
into linear terms exists in the "ring of objects"...
Maxim Stepin
"Arturo Magidin" <mag...@math.berkeley.edu> wrote in message
news:bhgnpr$144u$1...@agate.berkeley.edu...
> (I note he corrected 'set' to 'commutative ring',
> finally).
At least he's making progress.
His "Object Mathematics" page looks much more clear now.
Seems like he's trying to be more careful with definitions, that's a good thing.
Arturo Magidin
Maxim Stepin <maxim...@worldnet.att.net> wrote:
>
>"Arturo Magidin" <mag...@math.berkeley.edu> wrote in message
>news:bhgnpr$144u$1...@agate.berkeley.edu...
>
>> (I note he corrected 'set' to 'commutative ring',
>> finally).
>
>At least he's making progress.
>
>His "Object Mathematics" page looks much more clear now.
It would take a lot of work to make it less clear than it was...
However, in addition to the problems still existing with his main
definitions, you still have beauties like:
The base set is {-1, 1}.
Operators act on objects or sets of objects to give as a result an
object or set of objects when in a complete expression. The base
operations for a ring are addition and multiplication.
An expression is said to be complete if it results in an object or set
of objects, which is why the system is object oriented.
An expression is said to be intermediate, if it results in an
operator, or an operator - object pair.
The primary ring is the ring of integers, which is also the base ring.
Factor ring: Set made up of the non-integer objects resulting from
the factorization of a polynomial with integer coefficients, for which
no rational roots exist. The order of the factor ring is the degree
of the polynomial.
I have three basic operations, which are the operations of addition,
multiplication, and something I've called extraction.
[No definition of "extraction", however]
For instance 1/2 is an operator, which returns a member of the set of
objects that multiply times themselves to give another object.
So, 11/2=(1/2){-1,1} = {1}, or {-1}, as a set operation.
Theorem of Objects: A factorization exists for any ring bridge where
all coefficients are objects for which the resultant ring bridge
factors only contain objects.
Rings can be crossed using ring bridges, which at their simplest are
polynomial expressions.
The separator x is the base of the ring bridge.
Also a number is an object if it is a factor of a polynomial
expression with object coefficients.
[Thus expanding the notion of object beyond the original definition;
now "objects" can apparently also be functions!]
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Tralfaz
> In article <bhhtkl$sfa6$1...@ID-57378.news.uni-berlin.de>,
> Maxim Stepin <maxim...@worldnet.att.net> wrote:
> >
> >"Arturo Magidin" <mag...@math.berkeley.edu> wrote in message
> >news:bhgnpr$144u$1...@agate.berkeley.edu...
> >
> >> (I note he corrected 'set' to 'commutative ring',
> >> finally).
> >
> >At least he's making progress.
> >
> >His "Object Mathematics" page looks much more clear now.
>
> It would take a lot of work to make it less clear than it was...
>
> However, in addition to the problems still existing with his main
> definitions, you still have beauties like:
>
>
> The base set is {-1, 1}.
Has anyone else noticed that according to the definition, no non-integral
rational number is an object e.g. 7/19 or 42/13
Arturo Magidin
>
>"Arturo Magidin" <mag...@math.berkeley.edu> wrote in message
>news:bhjeub$2tjh$1...@agate.berkeley.edu...
>> In article <bhhtkl$sfa6$1...@ID-57378.news.uni-berlin.de>,
>> Maxim Stepin <maxim...@worldnet.att.net> wrote:
>> >
>> >"Arturo Magidin" <mag...@math.berkeley.edu> wrote in message
>> >news:bhgnpr$144u$1...@agate.berkeley.edu...
>> >
>> >> (I note he corrected 'set' to 'commutative ring',
>> >> finally).
>> >
>> >At least he's making progress.
>> >
>> >His "Object Mathematics" page looks much more clear now.
>>
>> It would take a lot of work to make it less clear than it was...
>>
>> However, in addition to the problems still existing with his main
>> definitions, you still have beauties like:
>>
>>
>> The base set is {-1, 1}.
>
>Has anyone else noticed that according to the definition, no non-integral
>rational number is an object e.g. 7/19 or 42/13
Which definition?
Yes, since it states that the only integers that are units are 1 and
-1, there cannot be any non-integral rational. Bill Dubuque pointed
out that this condition, and the unstated condition James implies
every once in a while of closure under Galois conjugates, implies that
this object ring must be a subring of the ring of all algebraic
integers (assuming he wants to stay within the algebraic numbers,
anyway).
Arturo Magidin
Maxim Stepin <maxim...@worldnet.att.net> wrote:
>
Well, he's regressing again:
"The Object Ring is a commutative ring that includes all numbers such
that -1 and 1 are the only members that are both a unit and an
integer, where no non-unit member is a factor of any two integers that
are coprime.
"A commutative ring is a set of numbers where for any members a and b,
a+b = b+a and ab = ba."
Every subset of a commutative ring is now a commutative ring. In fact,
any set is a commutative ring: if empty, it is a ring by vacuity; if
nonempty, let x_0 in S, and define +:SxS -> S and *:SxS -> S to be
given by
a+b = x_0
a*b = x_0
for all a,b in S. Voila! It's a "commutative ring". And I even made
sure the set was closed under the operations. James's "definition"
does not even require ->that<-.
HINT TO JAMES:
Here's what the definition of "commutative ring" should be:
DEF. A "commutative ring" is a set R, together with two binary
operations (that is, functions from RxR to R), usually denoted by "+"
and "*" and called "addition" and "multiplication", respectively; we
also usually use infix notation on these operators. The set R and the
operations + and * must satisfy the following:
(i) For every a,b,c in R, (a+b)+c = a+(b+c).
(ii) There exists an element of R, called 0, such that for all a in R
a+0=0+a = a.
(iii) For every a in R there exists b in R (which depends on a) such
that a+b=b+a=0.
(iv) For every a,b in R, a+b=b+a.
(v) For every a,b,c in R, (a*b)*c = a*(b*c).
(vi) For every a,b,c in R, a*(b+c) = (a*b)+(a*c),
and (b+c)*a = (b*a)+(c*a).
(vii) For every a,b in R, a*b=b*a.
If a set satisfies only (i)-(vi), it is called a "ring" (or
non-commutative ring, for emphasis).
If the ring also satisfies:
(viii) There exists an element of R, called "1", such that for all a
in R a*1=1*a=a.
then we say that R is a "ring with identity" or "ring with 1" (if it
is not necessarily commutative), or a "commutative ring with 1" or
"commutative ring with identity" if it satisfies (vii).
If a commutative ring with 1 also satisfies:
(ix) For every a,b in R, if a*b=0 then a=0 or b=0;
then the ring is said to be a "Domain" or an "integral domain". The
condition (ix) is easily proven to be equivalent to:
(ix') For every a,b,c in R, if a*b=a*c and a is not equal to 0, then
b=c.